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Double-Angle, Half-Angle, and Reduction Formulas

Use double-angle formulas to find exact values.

Using Double-Angle Formulas to Find Exact Values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where \(\alpha =\beta .\) Deriving the double-angle formula for sine begins with the sum formula,

\[\sin (\alpha +\beta )=\sin \ \alpha \ \cos \ \beta +\cos \ \alpha \ \sin \ \beta\]

If we let \(\alpha =\beta =\theta ,\) then we have

\[\begin{array}{lll}\sin (\theta +\theta ) & = & \sin \ \theta \ \cos \ \theta +\cos \ \theta \ \sin \ \theta \\ \sin (2\theta ) & = & 2\sin \ \theta \ \cos \ \theta \end{array}\]

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, \(\cos (\alpha +\beta )=\cos \ \alpha \ \cos \ \beta -\sin \ \alpha \ \sin \ \beta ,\) and letting \(\alpha =\beta =\theta ,\) we have

\[\begin{array}{lll}\cos (\theta +\theta ) & = & \cos \ \theta \ \cos \ \theta -\sin \ \theta \ \sin \ \theta \\ \cos (2\theta ) & = & {\cos }^{2}\theta -{\sin }^{2}\theta \end{array}\]

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more variations. The first variation is:

\[\begin{array}{lll}\cos (2\theta ) & = & {\cos }^{2}\theta -{\sin }^{2}\theta \\ & = & (1-{\sin }^{2}\theta )-{\sin }^{2}\theta \\ & = & 1-2{\sin }^{2}\theta \end{array}\]

The second variation is:

\[\begin{array}{lll}\cos (2\theta ) & = & {\cos }^{2}\theta -{\sin }^{2}\theta \\ & = & {\cos }^{2}\theta -(1-{\cos }^{2}\theta ) \\ & = & 2\ {\cos }^{2}\theta -1\end{array}\]

Similarly, to derive the double-angle formula for tangent, replacing \(\alpha =\beta =\theta\) in the sum formula gives

\[\begin{array}{lll}\tan (\alpha +\beta ) & = & \frac{\tan \ \alpha +\tan \ \beta }{1-\tan \ \alpha \ \tan \ \beta } \\ \tan (\theta +\theta ) & = & \frac{\tan \ \theta +\tan \ \theta }{1-\tan \ \theta \ \tan \ \theta } \\ \tan (2\theta ) & = & \frac{2\tan \ \theta }{1-{\tan }^{2}\theta }\end{array}\]

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Using Double-Angle Formulas to Verify Identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Example

Try it.

Verify the following identity using double-angle formulas:

\[1+\sin (2\theta )={(\sin \ \theta +\cos \ \theta )}^{2}\]

Solution

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

\[\begin{array}{lll}{(\sin \ \theta +\cos \ \theta )}^{2} & = & {\sin }^{2}\theta +2\ \sin \ \theta \ \cos \ \theta +{\cos }^{2}\theta \\ & = & ({\sin }^{2}\theta +{\cos }^{2}\theta )+2\ \sin \ \theta \ \cos \ \theta \\ & = & 1+2\ \sin \ \theta \ \cos \ \theta \\ & = & 1+\sin (2\theta )\end{array}\]

Example

Try it.

Verify the identity:

\[\tan (2\theta )=\frac{2}{\text{cot}\ \theta -\tan \ \theta }\]

Solution

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

\[\begin{array}{llll}\tan (2\theta ) & = & \frac{2\ \tan \ \theta }{1-{\tan }^{2}\theta } & \ \text{Double-angle formula} \\ & = & \frac{2\ \tan \ \theta (\frac{1}{\tan \ \theta })}{(1-{\tan }^{2}\theta )(\frac{1}{\tan \ \theta })} & \ \text{Multiply by a term that results in desired numerator}. \\ & = & \frac{2}{\frac{1}{\tan \ \theta }-\frac{{\tan }^{2}\theta }{\tan \ \theta }} & \\ & = & \frac{2}{\text{cot}\ \theta -\tan \ \theta } & \ \text{Use reciprocal identity for }\frac{1}{\tan \ \theta }.\end{array}\]

Use Reduction Formulas to Simplify an Expression

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with \(\cos (2\theta )=1-2\ {\sin }^{2}\theta .\) Solve for \({\sin }^{2}\theta :\)

\[\begin{array}{lll}\cos (2\theta ) & = & 1-2\ {\sin }^{2}\theta \\ 2\ {\sin }^{2}\theta & = & 1-\cos (2\theta ) \\ {\sin }^{2}\theta & = & \frac{1-\cos (2\theta )}{2}\end{array}\]

Next, we use the formula \(\cos (2\theta )=2\ {\cos }^{2}\theta -1.\) Solve for \({\cos }^{2}\theta :\)

\[\begin{array}{lll}\cos (2\theta ) & = & \ 2\ {\cos }^{2}\theta -1 \\ 1+\cos (2\theta ) & = & 2\ {\cos }^{2}\theta \\ \frac{1+\cos (2\theta )}{2} & = & {\cos }^{2}\theta \end{array}\]

The last reduction formula is derived by writing tangent in terms of sine and cosine:

\[\begin{array}{llll}{\tan }^{2}\theta & = & \frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } & \\ & = & \frac{\frac{1-\cos (2\theta )}{2}}{\frac{1+\cos (2\theta )}{2}} & \ \text{Substitute the reduction formulas}\text{.} \\ & = & (\frac{1-\cos (2\theta )}{2})(\frac{2}{1+\cos (2\theta )}) & \\ & = & \frac{1-\cos (2\theta )}{1+\cos (2\theta )} & \end{array}\]

Example

Try it.

Write an equivalent expression for \({\cos }^{4}x\) that does not involve any powers of sine or cosine greater than 1.

Solution

We will apply the reduction formula for cosine twice.

\[\begin{array}{llll}{\cos }^{4}x & = & {({\cos }^{2}x)}^{2} & \\ & = & {(\frac{1+\cos (2x)}{2})}^{2} & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{1}{4}(1+2\cos (2x)+{\cos }^{2}(2x)) & \\ & = & \frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{4}(\frac{1+\cos 2(2x)}{2}) & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}+\frac{1}{8}\ \cos (4x) & \\ & = & \frac{3}{8}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}\ \cos (4x) & \end{array}\]

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace \(\theta\) with \(\frac{\alpha }{2},\) the half-angle formula for sine is found by simplifying the equation and solving for \(\sin (\frac{\alpha }{2}).\) Note that the half-angle formulas are preceded by a \(\pm\) sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which \(\frac{\alpha }{2}\) terminates.

The half-angle formula for sine is derived as follows:

\[\begin{array}{lll}{\sin }^{2}\theta & = & \frac{1-\cos (2\theta )}{2} \\ {\sin }^{2}(\frac{\alpha }{2}) & = & \frac{1-(\cos 2⋅\frac{\alpha }{2})}{2} \\ & = & \frac{1-\cos \ \alpha }{2} \\ \sin (\frac{\alpha }{2}) & = & \pm \sqrt{\frac{1-\cos \ \alpha }{2}}\end{array}\]

To derive the half-angle formula for cosine, we have

\[\begin{array}{lll}{\cos }^{2}\theta & = & \frac{1+\cos (2\theta )}{2} \\ {\cos }^{2}(\frac{\alpha }{2}) & = & \frac{1+\cos (2⋅\frac{\alpha }{2})}{2} \\ & = & \frac{1+\cos \ \alpha }{2} \\ \cos (\frac{\alpha }{2}) & = & \pm \sqrt{\frac{1+\cos \ \alpha }{2}}\end{array}\]

For the tangent identity, we have

\[\begin{array}{lll}{\tan }^{2}\theta & = & \frac{1-\cos (2\theta )}{1+\cos (2\theta )} \\ {\tan }^{2}(\frac{\alpha }{2}) & = & \frac{1-\cos (2⋅\frac{\alpha }{2})}{1+\cos (2⋅\frac{\alpha }{2})} \\ & = & \frac{1-\cos \ \alpha }{1+\cos \ \alpha } \\ \tan (\frac{\alpha }{2}) & = & \pm \sqrt{\frac{1-\cos \ \alpha }{1+\cos \ \alpha }}\end{array}\]

Example

Try it.

Find \(\sin (15^{\circ})\) using a half-angle formula.

Solution

Since \(15^{\circ}=\frac{30^{\circ}}{2},\) we use the half-angle formula for sine:

\[\begin{array}{lll}\sin \ \frac{30^{\circ}}{2} & = & \sqrt{\frac{1-\cos 30^{\circ}}{2}} \\ & = & \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ & = & \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}} \\ & = & \sqrt{\frac{2-\sqrt{3}}{4}} \\ & = & \frac{\sqrt{2-\sqrt{3}}}{2}\end{array}\]

Remember that we can check the answer with a graphing calculator.

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Key Equations

Double-angle formulas	\(\begin{array}{lll}\sin (2\theta ) & = & 2\sin \ \theta \ \cos \ \theta \\ \cos (2\theta ) & = & {\cos }^{2}\theta -{\sin }^{2}\theta \\ & = & 1-2{\sin }^{2}\theta \\ & = & 2{\cos }^{2}\theta -1 \\ \tan (2\theta ) & = & \frac{2\tan \ \theta }{1-{\tan }^{2}\theta }\end{array}\)
Reduction formulas	\(\begin{array}{lll}{\sin }^{2}\theta & = & \frac{1-\cos (2\theta )}{2} \\ {\cos }^{2}\theta & = & \frac{1+\cos (2\theta )}{2} \\ {\tan }^{2}\theta & = & \frac{1-\cos (2\theta )}{1+\cos (2\theta )}\end{array}\)
Half-angle formulas	\(\begin{array}{lll}\sin \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1-\cos \ \alpha }{2}} \\ \cos \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1+\cos \ \alpha }{2}} \\ \tan \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1-\cos \ \alpha }{1+\cos \ \alpha }} \\ & = & \frac{\sin \ \alpha }{1+\cos \ \alpha } \\ & = & \frac{1-\cos \ \alpha }{\sin \ \alpha }\end{array}\)

Key Concepts

Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See , , , and .
Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See and .
Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See , , and .

Using Double-Angle Formulas to Find Exact Values

\[\sin (\alpha +\beta )=\sin \ \alpha \ \cos \ \beta +\cos \ \alpha \ \sin \ \beta\]

If we let \(\alpha =\beta =\theta ,\) then we have

\[\begin{array}{l}\sin (\theta +\theta )=\sin \ \theta \ \cos \ \theta +\cos \ \theta \ \sin \ \theta \\ \sin (2\theta )=2\sin \ \theta \ \cos \ \theta \end{array}\]

\[\begin{array}{l}\cos (\theta +\theta )=\cos \ \theta \ \cos \ \theta -\sin \ \theta \sin \ \theta \\ \cos (2\theta )={\cos }^{2}\theta -{\sin }^{2}\theta \end{array}\]

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

\[\begin{array}{l}\cos (2\theta )={\cos }^{2}\theta -{\sin }^{2}\theta \\ =(1-{\sin }^{2}\theta )-{\sin }^{2}\theta \\ =1-2{\sin }^{2}\theta \end{array}\]

The second interpretation is:

\[\begin{array}{l}\cos (2\theta )={\cos }^{2}\theta -{\sin }^{2}\theta \\ ={\cos }^{2}\theta -(1-{\cos }^{2}\theta ) \\ =2\ {\cos }^{2}\theta -1\end{array}\]

Similarly, to derive the double-angle formula for tangent, replacing \(\alpha =\beta =\theta\) in the sum formula gives

\[\begin{array}{l}\tan (\alpha +\beta )=\frac{\tan \ \alpha +\tan \ \beta }{1-\tan \ \alpha \ \tan \ \beta } \\ \tan (\theta +\theta )=\frac{\tan \ \theta +\tan \ \theta }{1-\tan \ \theta \ \tan \ \theta } \\ \tan (2\theta )=\frac{2\tan \ \theta }{1-{\tan }^{2}\theta }\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Using Double-Angle Formulas to Verify Identities

Example

Try it.

Establish the following identity using double-angle formulas:

\[1+\sin (2\theta )={(\sin \ \theta +\cos \ \theta )}^{2}\]

Solution

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

\[\begin{array}{l}{(\sin \ \theta +\cos \ \theta )}^{2}={\sin }^{2}\theta +2\ \sin \ \theta \ \cos \ \theta +{\cos }^{2}\theta \\ =({\sin }^{2}\theta +{\cos }^{2}\theta )+2\ \sin \ \theta \ \cos \ \theta \\ =1+2\ \sin \ \theta \ \cos \ \theta \\ =1+\sin (2\theta )\end{array}\]

Example

Try it.

Verify the identity:

\[\tan (2\theta )=\frac{2}{\text{cot}\ \theta -\tan \ \theta }\]

Solution

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

\[\begin{array}{lllll}\tan (2\theta )=\frac{2\ \tan \ \theta }{1-{\tan }^{2}\theta } & \text{Double-angle formula} \\ =\frac{2\ \tan \ \theta (\frac{1}{\tan \ \theta })}{(1-{\tan }^{2}\theta )(\frac{1}{\tan \ \theta })}\begin{array}{llll} & & & \end{array} & \text{Multiply by a term that results in desired numerator}. \\ =\frac{2}{\frac{1}{\tan \ \theta }-\frac{{\tan }^{2}\theta }{\tan \ \theta }} & \\ =\frac{2}{\text{cot}\ \theta -\tan \ \theta } & \text{Use reciprocal identity for }\frac{1}{\tan \ \theta }.\end{array}\]

Use Reduction Formulas to Simplify an Expression

\[\begin{array}{l}\cos (2\theta )=1-2\ {\sin }^{2}\theta \\ 2\ {\sin }^{2}\theta =1-\cos (2\theta ) \\ {\sin }^{2}\theta =\frac{1-\cos (2\theta )}{2}\end{array}\]

Next, we use the formula \(\cos (2\theta )=2\ {\cos }^{2}\theta -1.\) Solve for \({\cos }^{2}\theta :\)

\[\begin{array}{l}\ \cos (2\theta )=2\ {\cos }^{2}\theta -1 \\ 1+\cos (2\theta )=2\ {\cos }^{2}\theta \\ \frac{1+\cos (2\theta )}{2}={\cos }^{2}\theta \end{array}\]

The last reduction formula is derived by writing tangent in terms of sine and cosine:

\[\begin{array}{lllll}{\tan }^{2}\theta =\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } & \\ =\frac{\frac{1-\cos (2\theta )}{2}}{\frac{1+\cos (2\theta )}{2}} & \text{Substitute the reduction formulas.} \\ =(\frac{1-\cos (2\theta )}{2})(\frac{2}{1+\cos (2\theta )})\begin{array}{llll} & & & \end{array} & \\ =\frac{1-\cos (2\theta )}{1+\cos (2\theta )} & \end{array}\]

Example

Try it.

Write an equivalent expression for \({\cos }^{4}x\) that does not involve any powers of sine or cosine greater than 1.

Solution

We will apply the reduction formula for cosine twice.

\[\begin{array}{lllll}{\cos }^{4}x={({\cos }^{2}x)}^{2} & \\ ={(\frac{1+\cos (2x)}{2})}^{2} & {\text{Substitute reduction formula for cos}}^{2}x. \\ =\frac{1}{4}(1+2\cos (2x)+{\cos }^{2}(2x)) & \\ =\frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{4}(\frac{1+\cos 2(2x)}{2})\begin{array}{llll} & & & \end{array} & {\text{ Substitute reduction formula for cos}}^{2}x. \\ =\frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}+\frac{1}{8}\ \cos (4x) & \\ =\frac{3}{8}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}\ \cos (4x) & \end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Using Half-Angle Formulas to Find Exact Values

The half-angle formula for sine is derived as follows:

\[\begin{array}{l}\ {\sin }^{2}\theta =\frac{1-\cos (2\theta )}{2} \\ {\sin }^{2}(\frac{\alpha }{2})=\frac{1-\cos (2⋅\frac{\alpha }{2})}{2} \\ =\frac{1-\cos \ \alpha }{2} \\ \sin (\frac{\alpha }{2})=\pm \sqrt{\frac{1-\cos \ \alpha }{2}}\end{array}\]

To derive the half-angle formula for cosine, we have

\[\begin{array}{l}\ {\cos }^{2}\theta =\frac{1+\cos (2\theta )}{2} \\ {\cos }^{2}(\frac{\alpha }{2})=\frac{1+\cos (2⋅\frac{\alpha }{2})}{2} \\ =\frac{1+\cos \ \alpha }{2} \\ \cos (\frac{\alpha }{2})=\pm \sqrt{\frac{1+\cos \ \alpha }{2}}\end{array}\]

For the tangent identity, we have

\[\begin{array}{l}\ {\tan }^{2}\theta =\frac{1-\cos (2\theta )}{1+\cos (2\theta )} \\ {\tan }^{2}(\frac{\alpha }{2})=\frac{1-\cos (2⋅\frac{\alpha }{2})}{1+\cos (2⋅\frac{\alpha }{2})} \\ =\frac{1-\cos \ \alpha }{1+\cos \ \alpha } \\ \tan (\frac{\alpha }{2})=\pm \sqrt{\frac{1-\cos \ \alpha }{1+\cos \ \alpha }}\end{array}\]

Example

Try it.

Find \(\sin ({15}^{∘})\) using a half-angle formula.

Solution

Since \({15}^{∘}=\frac{{30}^{∘}}{2},\) we use the half-angle formula for sine:

\[\begin{array}{l}\begin{array}{l} \\ \sin \ \frac{{30}^{∘}}{2}=\sqrt{\frac{1-\cos {30}^{∘}}{2}}\end{array} \\ =\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ =\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}} \\ =\sqrt{\frac{2-\sqrt{3}}{4}} \\ =\frac{\sqrt{2-\sqrt{3}}}{2}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

Double-angle formulas	\(\begin{array}{l}\sin (2\theta )=2\sin \ \theta \ \cos \ \theta \\ \cos (2\theta )={\cos }^{2}\theta -{\sin }^{2}\theta \\ =1-2{\sin }^{2}\theta \\ =2{\cos }^{2}\theta -1 \\ \tan (2\theta )=\frac{2\tan \ \theta }{1-{\tan }^{2}\theta }\end{array}\)
Reduction formulas	\(\begin{array}{l}{\sin }^{2}\theta =\frac{1-\cos (2\theta )}{2} \\ {\cos }^{2}\theta =\frac{1+\cos (2\theta )}{2} \\ {\tan }^{2}\theta =\frac{1-\cos (2\theta )}{1+\cos (2\theta )}\end{array}\)
Half-angle formulas	\(\begin{array}{l}\sin \ \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \ \alpha }{2}} \\ \cos \ \frac{\alpha }{2}=\pm \sqrt{\frac{1+\cos \ \alpha }{2}} \\ \tan \ \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \ \alpha }{1+\cos \ \alpha }} \\ =\frac{\sin \ \alpha }{1+\cos \ \alpha } \\ =\frac{1-\cos \ \alpha }{\sin \ \alpha }\end{array}\)

Key Concepts

Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See , , , and .
Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See and .
Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See , , and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Given that \(\tan \ \theta =-\frac{3}{4}\) and \(\theta\) is in quadrant II, find the following:
1. ⓐ \(\sin (2\theta )\)
2. ⓑ \(\cos (2\theta )\)
3. ⓒ \(\tan (2\theta )\)
Хариулт
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given \(\tan \ \theta =-\frac{3}{4},\) such that \(\theta\) is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because \(\theta\) is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
\[\begin{array}{lll}{(-4)}^{2}+{(3)}^{2} & = & {c}^{2} \\ 16+9 & = & {c}^{2} \\ 25 & = & {c}^{2} \\ c & = & 5\end{array}\]
Now we can draw a triangle similar to the one shown in .
1. ⓐ Let’s begin by writing the double-angle formula for sine. \[\sin (2\theta )=2\ \sin \ \theta \ \cos \ \theta\]
  We see that we to need to find \(\sin \ \theta\) and \(\cos \ \theta .\) Based on , we see that the hypotenuse equals 5, so \(\sin \ \theta =\frac{3}{5},\) and \(\cos \ \theta =-\frac{4}{5}.\) Substitute these values into the equation, and simplify.
  Thus,
  \[\begin{array}{lll}\sin (2\theta ) & = & 2(\frac{3}{5})(-\frac{4}{5}) \\ & = & -\frac{24}{25}\end{array}\]
2. ⓑ Write the double-angle formula for cosine. \[\cos (2\theta )={\cos }^{2}\theta -{\sin }^{2}\theta\]
  Again, substitute the values of the sine and cosine into the equation, and simplify.
  \[\begin{array}{lll}\cos (2\theta ) & = & {(-\frac{4}{5})}^{2}-{(\frac{3}{5})}^{2} \\ & = & \frac{16}{25}-\frac{9}{25} \\ & = & \frac{7}{25}\end{array}\]
3. ⓒ Write the double-angle formula for tangent. \[\tan (2\theta )=\frac{2\ \tan \ \theta }{1-{\tan }^{2}\theta }\]
  In this formula, we need the tangent, which we were given as \(\tan \ \theta =-\frac{3}{4}.\) Substitute this value into the equation, and simplify.
  \[\begin{array}{lll}\tan (2\theta ) & = & \frac{2(-\frac{3}{4})}{1-{(-\frac{3}{4})}^{2}} \\ & = & \frac{-\frac{3}{2}}{1-\frac{9}{16}} \\ & = & -\frac{3}{2}(\frac{16}{7}) \\ & = & -\frac{24}{7}\end{array}\]
Given \(\sin \ \alpha =\frac{5}{8},\) with \(\alpha\) in quadrant I, find \(\cos (2\alpha ).\)

Хариулт

\(\cos (2\alpha )=\frac{7}{32}\)
Use the double-angle formula for cosine to write \(\cos (6x)\) in terms of \(\cos (3x).\)

Хариулт

\[\begin{array}{lll}\cos (6x) & = & \cos (2(3x)) \\ & = & {2\cos }^{2}(3x)-1\end{array}\]
Verify the following identity using double-angle formulas:
\[1+\sin (2\theta )={(\sin \ \theta +\cos \ \theta )}^{2}\]

Хариулт

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.
\[\begin{array}{lll}{(\sin \ \theta +\cos \ \theta )}^{2} & = & {\sin }^{2}\theta +2\ \sin \ \theta \ \cos \ \theta +{\cos }^{2}\theta \\ & = & ({\sin }^{2}\theta +{\cos }^{2}\theta )+2\ \sin \ \theta \ \cos \ \theta \\ & = & 1+2\ \sin \ \theta \ \cos \ \theta \\ & = & 1+\sin (2\theta )\end{array}\]
Verify the identity: \({\cos }^{4}\theta -{\sin }^{4}\theta =\cos (2\theta ).\)

Хариулт

\({\cos }^{4}\theta -{\sin }^{4}\theta =({\cos }^{2}\theta +{\sin }^{2}\theta )({\cos }^{2}\theta -{\sin }^{2}\theta )=\cos (2\theta )\)
Verify the identity:
\[\tan (2\theta )=\frac{2}{\text{cot}\ \theta -\tan \ \theta }\]

Хариулт

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.
\[\begin{array}{llll}\tan (2\theta ) & = & \frac{2\ \tan \ \theta }{1-{\tan }^{2}\theta } & \ \text{Double-angle formula} \\ & = & \frac{2\ \tan \ \theta (\frac{1}{\tan \ \theta })}{(1-{\tan }^{2}\theta )(\frac{1}{\tan \ \theta })} & \ \text{Multiply by a term that results in desired numerator}. \\ & = & \frac{2}{\frac{1}{\tan \ \theta }-\frac{{\tan }^{2}\theta }{\tan \ \theta }} & \\ & = & \frac{2}{\text{cot}\ \theta -\tan \ \theta } & \ \text{Use reciprocal identity for }\frac{1}{\tan \ \theta }.\end{array}\]
Verify the identity: \(\cos (2\theta )\cos \ \theta ={\cos }^{3}\theta -\cos \ \theta {\sin }^{2}\theta .\)

Хариулт

\(\cos (2\theta )\cos \ \theta =({\cos }^{2}\theta -{\sin }^{2}\theta )\cos \ \theta ={\cos }^{3}\theta -\cos \ \theta {\sin }^{2}\theta\)
Write an equivalent expression for \({\cos }^{4}x\) that does not involve any powers of sine or cosine greater than 1.

Хариулт

We will apply the reduction formula for cosine twice.
\[\begin{array}{llll}{\cos }^{4}x & = & {({\cos }^{2}x)}^{2} & \\ & = & {(\frac{1+\cos (2x)}{2})}^{2} & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{1}{4}(1+2\cos (2x)+{\cos }^{2}(2x)) & \\ & = & \frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{4}(\frac{1+\cos 2(2x)}{2}) & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{1}{4}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}+\frac{1}{8}\ \cos (4x) & \\ & = & \frac{3}{8}+\frac{1}{2}\ \cos (2x)+\frac{1}{8}\ \cos (4x) & \end{array}\]
Use the power-reducing formulas to prove
\[{\sin }^{3}(2x)=[\frac{1}{2}\ \sin (2x)]\ [1-\cos (4x)]\]

Хариулт

We will work on simplifying the left side of the equation:
\[\begin{array}{llll}{\sin }^{3}(2x) & = & [\sin (2x)][{\sin }^{2}(2x)] & \\ & = & \sin (2x)[\frac{1-\cos (4x)}{2}] & \text{Substitute the power-reduction formula}. \\ & = & \sin (2x)(\frac{1}{2})[1-\cos (4x)] & \\ & = & \frac{1}{2}[\sin (2x)][1-\cos (4x)] & \end{array}\]
Use the power-reducing formulas to prove that \(10\ {\cos }^{4}x=\frac{15}{4}+5\ \cos (2x)+\frac{5}{4}\ \cos (4x).\)

Хариулт

\(\begin{array}{llll}10{\cos }^{4}x & = & 10{({\cos }^{2}x)}^{2} & \\ & = & 10{[\frac{1+\cos (2x)}{2}]}^{2} & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{10}{4}[1+2\cos (2x)+{\cos }^{2}(2x)] & \\ & = & \frac{10}{4}+\frac{10}{2}\cos (2x)+\frac{10}{4}(\frac{1+\cos 2(2x)}{2}) & {\ \text{Substitute reduction formula for cos}}^{2}x. \\ & = & \frac{10}{4}+\frac{10}{2}\cos (2x)+\frac{10}{8}+\frac{10}{8}\cos (4x) & \\ & = & \frac{30}{8}+5\cos (2x)+\frac{10}{8}\cos (4x) & \\ & = & \frac{15}{4}+5\cos (2x)+\frac{5}{4}\cos (4x) & \end{array}\)
Find \(\sin (15^{\circ})\) using a half-angle formula.

Хариулт

Since \(15^{\circ}=\frac{30^{\circ}}{2},\) we use the half-angle formula for sine:
\[\begin{array}{lll}\sin \ \frac{30^{\circ}}{2} & = & \sqrt{\frac{1-\cos 30^{\circ}}{2}} \\ & = & \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ & = & \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}} \\ & = & \sqrt{\frac{2-\sqrt{3}}{4}} \\ & = & \frac{\sqrt{2-\sqrt{3}}}{2}\end{array}\]
Remember that we can check the answer with a graphing calculator.
Given that \(\tan \ \alpha =\frac{8}{15}\) and \(\alpha\) lies in quadrant III, find the exact value of the following:
1. ⓐ \(\sin (\frac{\alpha }{2})\)
2. ⓑ \(\cos (\frac{\alpha }{2})\)
3. ⓒ \(\tan (\frac{\alpha }{2})\)
Хариулт
Using the given information, we can draw the triangle shown in . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate \(\sin \ \alpha =-\frac{8}{17}\) and \(\cos \ \alpha =-\frac{15}{17}.\)
1. ⓐ Before we start, we must remember that if \(\alpha\) is in quadrant III, then \(180^{\circ}<\alpha <270^{\circ},\) so \(\frac{180^{\circ}}{2}<\frac{\alpha }{2}<\frac{270^{\circ}}{2}.\) This means that the terminal side of \(\frac{\alpha }{2}\) is in quadrant II, since \(90^{\circ}<\frac{\alpha }{2}<135^{\circ}.\)
  To find \(\sin \ \frac{\alpha }{2},\) we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in and simplify.
  \[\begin{array}{lll}\sin \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1-\cos \ \alpha }{2}} \\ & = & \pm \sqrt{\frac{1-(-\frac{15}{17})}{2}} \\ & = & \pm \sqrt{\frac{\frac{32}{17}}{2}} \\ & = & \pm \sqrt{\frac{32}{17}⋅\frac{1}{2}} \\ & = & \pm \sqrt{\frac{16}{17}} \\ & = & \pm \frac{4}{\sqrt{17}} \\ & = & \frac{4\sqrt{17}}{17}\end{array}\]
  We choose the positive value of \(\sin \ \frac{\alpha }{2}\) because the angle terminates in quadrant II and sine is positive in quadrant II.
2. ⓑ To find \(\cos \ \frac{\alpha }{2},\) we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in , and simplify. \[\begin{array}{lll}\cos \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1+\cos \ \alpha }{2}} \\ & = & \pm \sqrt{\frac{1+(-\frac{15}{17})}{2}} \\ & = & \pm \sqrt{\frac{\frac{2}{17}}{2}} \\ & = & \pm \sqrt{\frac{2}{17}⋅\frac{1}{2}} \\ & = & \pm \sqrt{\frac{1}{17}} \\ & = & -\frac{\sqrt{17}}{17}\end{array}\]
  We choose the negative value of \(\cos \ \frac{\alpha }{2}\) because the angle is in quadrant II because cosine is negative in quadrant II.
3. ⓒ To find \(\tan \ \frac{\alpha }{2},\) we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in and simplify. \[\begin{array}{lll}\tan \ \frac{\alpha }{2} & = & \pm \sqrt{\frac{1-\cos \ \alpha }{1+\cos \ \alpha }} \\ & = & \pm \sqrt{\frac{1-(-\frac{15}{17})}{1+(-\frac{15}{17})}} \\ & = & \pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}} \\ & = & \pm \sqrt{\frac{32}{2}} \\ & = & -\sqrt{16} \\ & = & -4\end{array}\]
  We choose the negative value of \(\tan \ \frac{\alpha }{2}\) because \(\frac{\alpha }{2}\) lies in quadrant II, and tangent is negative in quadrant II.
Given that \(\sin \ \alpha =-\frac{4}{5}\) and \(\alpha\) lies in quadrant IV, find the exact value of \(\cos \ (\frac{\alpha }{2}).\)

Хариулт

\(-\frac{2}{\sqrt{5}}\)
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of \(\theta\) formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If \(\tan \ \theta =\frac{5}{3}\) for higher-level competition, what is the measurement of the angle for novice competition?

Хариулт

Since the angle for novice competition measures half the steepness of the angle for the high level competition, and \(\tan \ \theta =\frac{5}{3}\) for high competition, we can find \(\cos \ \theta\) from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See .
\[\begin{array}{lll}{3}^{2}+{5}^{2} & = & 34 \\ c & = & \sqrt{34}\end{array}\]
We see that \(\cos \ \theta =\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}.\) We can use the half-angle formula for tangent: \(\tan \ \frac{\theta }{2}=\sqrt{\frac{1-\cos \ \theta }{1+\cos \ \theta }}.\) Since \(\tan \ \theta\) is in the first quadrant, so is \(\tan \ \frac{\theta }{2}.\)
\[\begin{array}{lll}\tan \ \frac{\theta }{2} & = & \sqrt{\frac{1-\frac{3\sqrt{34}}{34}}{1+\frac{3\sqrt{34}}{34}}} \\ & = & \sqrt{\frac{\frac{34-3\sqrt{34}}{34}}{\frac{34+3\sqrt{34}}{34}}} \\ & = & \sqrt{\frac{34-3\sqrt{34}}{34+3\sqrt{34}}} \\ & \approx & 0.57\end{array}\]
We can take the inverse tangent to find the angle: \({\tan }^{-1}(0.57)\approx 29.7^{\circ}.\) So the angle of the ramp for novice competition is \(\approx 29.7^{\circ}.\)
Explain how to determine the reduction identities from the double-angle identity \(\cos (2x)={\cos }^{2}x-{\sin }^{2}x.\)

Хариулт

Use the Pythagorean identities and isolate the squared term.
Explain how to determine the double-angle formula for \(\tan (2x)\) using the double-angle formulas for \(\cos (2x)\) and \(\sin (2x).\)
We can determine the half-angle formula for \(\tan (\frac{x}{2})=\frac{\sqrt{1-\cos \ x}}{\sqrt{1+\cos \ x}}\) by dividing the formula for \(\sin (\frac{x}{2})\) by \(\cos (\frac{x}{2}).\) Explain how to determine two formulas for \(\tan (\frac{x}{2})\) that do not involve any square roots.

Хариулт

\(\frac{1-\cos \ x}{\sin \ x},\frac{\sin \ x}{1+\cos \ x},\) multiplying the top and bottom by \(\sqrt{1-\cos \ x}\) and \(\sqrt{1+\cos \ x},\) respectively.
For the half-angle formula given in the previous exercise for \(\tan (\frac{x}{2}),\) explain why dividing by 0 is not a concern. (Hint: examine the values of \(\cos \ x\) necessary for the denominator to be 0.)
If \(\sin \ x=\frac{1}{8},\) and \(x\) is in quadrant I.

Хариулт

a) \(\frac{3\sqrt{7}}{32}\) b) \(\frac{31}{32}\) c) \(\frac{3\sqrt{7}}{31}\)
If \(\cos \ x=\frac{2}{3},\) and \(x\) is in quadrant I.
If \(\cos \ x=-\frac{1}{2},\) and \(x\\) is in quadrant III.

Хариулт

a) \(\frac{\sqrt{3}}{2}\) b) \(-\frac{1}{2}\) c) \(-\sqrt{3}\)
If \(\tan \ x=-8,\) and \(x\) is in quadrant IV.
\(\cos (2\theta )=\frac{3}{5}\) and \(90^{\circ}\le \theta \le 180^{\circ}\)

Хариулт

\(\cos \ \theta =-\frac{2\sqrt{5}}{5},\sin \ \theta =\frac{\sqrt{5}}{5},\tan \ \theta =-\frac{1}{2},\text{csc}\ \theta =\sqrt{5},\text{sec}\ \theta =-\frac{\sqrt{5}}{2},\text{cot}\ \theta =-2\)
\(\cos (2\theta )=\frac{1}{\sqrt{2}}\) and \(180^{\circ}\le \theta \le 270^{\circ}\)
\(2\ \sin (\frac{\pi }{4})\cos (\frac{\pi }{4})\)

Хариулт

\(\ \sin (\frac{\pi }{2})\)
\(4\ \sin (\frac{\pi }{8})\ \cos (\frac{\pi }{8})\)
\(\sin (\frac{\pi }{8})\\)

Хариулт

\(\frac{\sqrt{2-\sqrt{2}}}{2}\)
\(\cos (-\frac{11\pi }{12})\)
\(\sin (\frac{11\pi }{12})\)

Хариулт

\(\frac{\sqrt{2-\sqrt{3}}}{2}\)
\(\cos (\frac{7\pi }{8})\)
\(\tan (\frac{5\pi }{12})\)

Хариулт

\(2+\sqrt{3}\)
\(\tan (-\frac{3\pi }{12})\)
\(\tan (-\frac{3\pi }{8})\)

Хариулт

\(-1-\sqrt{2}\)
If \(\tan \ x=-\frac{4}{3},\) and \(x\) is in quadrant IV.
If \(\sin \ x=-\frac{12}{13},\) and \(x\) is in quadrant III.

Хариулт

a) \(\frac{3\sqrt{13}}{13}\) b) \(-\frac{2\sqrt{13}}{13}\) c) \(-\frac{3}{2}\)
If \(\text{csc}\ x=7,\) and \(\ x\\) is in quadrant II.
If \(\text{sec}\ x=-4,\) and \(x\) is in quadrant II.

Хариулт

a) \(\frac{\sqrt{10}}{4}\) b) \(\frac{\sqrt{6}}{4}\) c) \(\frac{\sqrt{15}}{3}\)
Find \(\sin (2\theta ),\cos (2\theta ),\) and \(\tan (2\theta ).\)
Find \(\sin (2\alpha ),\cos (2\alpha ),\) and \(\tan (2\alpha ).\)

Хариулт

\(\frac{120}{169},-\frac{119}{169},-\frac{120}{119}\)
Find \(\sin (\frac{\theta }{2}),\cos (\frac{\theta }{2}),\) and \(\tan (\frac{\theta }{2}).\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Double-Angle, Half-Angle, and Reduction Formulas

Use double-angle formulas to find exact values.
Use double-angle formulas to verify identities.
Use reduction formulas to simplify an expression.
Use half-angle formulas to find exact values.
Draw a triangle to reflect the given information.
Determine the correct double-angle formula.
Substitute values into the formula based on the triangle.
Simplify.

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Өөрийнхөөг турш

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Бүх зүйл Trigonometry

The unit circle Trigonometric equations Trigonometric identities Degrees and radians Right-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosines Graphs of sine, cosine and tangent Inverse trigonometric functions