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Trigonometric Functions

Convert angle measures between degrees and radians.

Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle \(\theta ,\) let \(s\) be the length of the corresponding arc on the unit circle (). We say the angle corresponding to the arc of length 1 has radian measure 1.

Since an angle of \(360\text{^{\circ}}\) corresponds to the circumference of a circle, or an arc of length \(2\pi ,\) we conclude that an angle with a degree measure of \(360\text{^{\circ}}\) has a radian measure of \(2\pi .\) Similarly, we see that \(180\text{^{\circ}}\) is equivalent to \(\pi\) radians. shows the relationship between common degree and radian values.

Degrees	Radians	Degrees	Radians
0	0	120	\(2\pi \text{/}3\)
30	\(\pi \text{/}6\)	135	\(3\pi \text{/}4\)
45	\(\pi \text{/}4\)	150	\(5\pi \text{/}6\)
60	\(\pi \text{/}3\)	180	\(\pi\)
90	\(\pi \text{/}2\)

Example

Try it.

Express \(225\text{^{\circ}}\) using radians.
Express \(5\pi \text{/}3\) rad using degrees.

Solution

Use the fact that \(180\text{^{\circ}}\) is equivalent to \(\pi\) radians as a conversion factor: \(1=\frac{\pi \ \text{rad}}{180\text{^{\circ}}}=\frac{180\text{^{\circ}}}{\pi \ \text{rad}}.\)

\(225\text{^{\circ}}=225\text{^{\circ}}\cdot \frac{\pi }{180\text{^{\circ}}}=\frac{5\pi }{4}\) rad
\(\frac{5\pi }{3}\) rad = \(\frac{5\pi }{3}\cdot \frac{180\text{^{\circ}}}{\pi }=300\text{^{\circ}}\)

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point \(P=(x,y)\) on the unit circle. Let \(\theta\) be an angle with an initial side that lies along the positive \(x\)-axis and with a terminal side that is the line segment \(OP.\) An angle in this position is said to be in standard position (). We can then define the values of the six trigonometric functions for \(\theta\) in terms of the coordinates \(x\) and \(y.\)

We can see that for a point \(P=(x,y)\) on a circle of radius \(r\) with a corresponding angle \(\theta ,\) the coordinates \(x\) and \(y\) satisfy

\[\begin{array}{l} \\ \\ \text{cos}\ \theta =\frac{x}{r} \\ x=r\ \text{cos}\ \theta \end{array}\]\[\begin{array}{l}\text{sin}\ \theta =\frac{y}{r} \\ y=r\ \text{sin}\ \theta .\end{array}\]

The values of the other trigonometric functions can be expressed in terms of \(x,y,\) and \(r\) ().

shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of \(\text{sin}\ \theta\) and \(\text{cos}\ \theta .\)

\(\theta\)	\(sin\ \theta\)	\(cos\ \theta\)
\(0\)	\(0\)	\(1\)
\(\frac{\pi }{6}\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)
\(\frac{\pi }{4}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{2}}{2}\)
\(\frac{\pi }{3}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)
\(\frac{\pi }{2}\)	\(1\)	\(0\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric functions that is true for all angles \(\theta\) for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

Rule: Trigonometric Identities

Reciprocal identities

\[\begin{array}{llll}\text{cot}\ \theta =\frac{1}{\text{tan}\ \theta } & & & \\ \text{csc}\ \theta =\frac{1}{\text{sin}\ \theta } & & & \text{sec}\ \theta =\frac{1}{\text{cos}\ \theta }\end{array}\]

Ratio identities

\[\begin{array}{llll}\text{tan}\ \theta =\frac{\text{sin}\ \theta }{\text{cos}\ \theta } & & & \text{cot}\ \theta =\frac{\text{cos}\ \theta }{\text{sin}\ \theta }\end{array}\]

Pythagorean identities

\[{\text{sin}}^{2}\theta +{\text{cos}}^{2}\theta =1\ 1+{\text{tan}}^{2}\theta ={\text{sec}}^{2}\theta \ 1+{\text{cot}}^{2}\theta ={\text{csc}}^{2}\theta\]

Addition and subtraction formulas

\[\text{sin}(\alpha \pm \beta )=\text{sin}\ \alpha \ \text{cos}\ \beta \pm \text{cos}\ \alpha \ \text{sin}\ \beta\]\[\text{cos}(\alpha \pm \beta )=\text{cos}\ \alpha \ \text{cos}\ \beta ∓\text{sin}\ \alpha \ \text{sin}\ \beta\]

Double-angle formulas

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

Example

Try it.

Prove the trigonometric identity \(1+{\text{tan}}^{2}\theta ={\text{sec}}^{2}\theta .\)

Solution

We start with the identity

\[{\text{sin}}^{2}\theta +{\text{cos}}^{2}\theta =1.\]

Dividing both sides of this equation by \({\text{cos}}^{2}\theta ,\) we obtain

\[\frac{{\text{sin}}^{2}\theta }{{\text{cos}}^{2}\theta }+1=\frac{1}{{\text{cos}}^{2}\theta }.\]

Since \(\text{sin}\ \theta \text{/}\text{cos}\ \theta =\text{tan}\ \theta\) and \(1\text{/}\text{cos}\ \theta =\text{sec}\ \theta ,\) we conclude that

\[{\text{tan}}^{2}\theta +1={\text{sec}}^{2}\theta .\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Graphs and Periods of the Trigonometric Functions

We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let \(P=(x,y)\) be a point on the unit circle and let \(\theta\) be the corresponding angle \(.\) Since the angle \(\theta\) and \(\theta +2\pi\) correspond to the same point \(P,\) the values of the trigonometric functions at \(\theta\) and at \(\theta +2\pi\) are the same. Consequently, the trigonometric functions are periodic functions. The period of a function \(f\) is defined to be the smallest positive value \(p\) such that \(f(x+p)=f(x)\) for all values \(x\) in the domain of \(f.\) The sine, cosine, secant, and cosecant functions have a period of \(2\pi .\) Since the tangent and cotangent functions repeat on an interval of length \(\pi ,\) their period is \(\pi\) ().

Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:

\[f(x)=A\ \text{cos}(B(x-\alpha ))+C.\]

In , the constant \(\alpha\) causes a horizontal or phase shift. The factor \(B\) changes the period. This transformed sine function will have a period \(2\pi \text{/}|B|.\) The factor \(A\) results in a vertical stretch by a factor of \(|A|.\) We say \(|A|\) is the “amplitude of \(f.\)” The constant \(C\) causes a vertical shift.

Notice in that the graph of \(y=\text{cos}\ x\) is the graph of \(y=\text{sin}\ x\) shifted to the left \(\pi \text{/}2\) units. Therefore, we can write \(\text{cos}\ x=\text{sin}(x+\pi \text{/}2).\) Similarly, we can view the graph of \(y=\text{sin}\ x\) as the graph of \(y=\text{cos}\ x\) shifted right \(\pi \text{/}2\) units, and state that \(\text{sin}\ x=\text{cos}(x-\pi \text{/}2).\)

A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with \(15.7\) hours and December 21 is the shortest day of the year with \(8.3\) hours. It can be shown that the function

\[h(t)=3.7\ \text{sin}(\frac{2\pi }{365}(t-80.5))+12\]

is a model for the number of hours of daylight \(h\) as a function of day of the year \(t\) ().

Example

Try it.

Sketch a graph of \(f(x)=3\ \text{sin}(2(x-\frac{\pi }{4}))+1.\)

Solution

This graph is a horizontal compression by a factor of 2, a phase shift to the right by π/4 units, followed by a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of \(f\) is \(\pi .\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of \(180\text{^{\circ}}\) has a radian measure of \(\pi\) rad.
For acute angles \(\theta ,\) the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is \(\theta .\)
For a general angle \(\theta ,\) let \((x,y)\) be a point on a circle of radius \(r\) corresponding to this angle \(\theta .\) The trigonometric functions can be written as ratios involving \(x,y,\) and \(r.\)
The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period \(2\pi .\) The tangent and cotangent functions have period \(\pi .\)

Trigonometric Functions

For the following exercises, convert each angle in degrees to radians. Write the answer as a multiple of \(\pi .\)

For the following exercises, convert each angle in radians to degrees.

Evaluate the following functional values.

For the following exercises, consider triangle ABC, a right triangle with a right angle at C. a. Find the missing side of the triangle. b. Find the six trigonometric function values for the angle at A. Where necessary, simplify to a fraction or round to three decimal places.

For the following exercises, \(P\) is a point on the unit circle. a. Find the (exact) missing coordinate value of each point and b. find the values of the six trigonometric functions for the angle \(\theta\) with a terminal side that passes through point \(P.\) Rationalize denominators.

For the following exercises, simplify each expression by writing it in terms of sines and cosines, then simplify. The final answer does not have to be in terms of sine and cosine only.

For the following exercises, verify that each equation is an identity.

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

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

1. Express \(225\text{^{\circ}}\) using radians.
2. Express \(5\pi \text{/}3\) rad using degrees.
જવાબ બતાવો
Use the fact that \(180\text{^{\circ}}\) is equivalent to \(\pi\) radians as a conversion factor: \(1=\frac{\pi \ \text{rad}}{180\text{^{\circ}}}=\frac{180\text{^{\circ}}}{\pi \ \text{rad}}.\)
1. \(225\text{^{\circ}}=225\text{^{\circ}}\cdot \frac{\pi }{180\text{^{\circ}}}=\frac{5\pi }{4}\) rad
2. \(\frac{5\pi }{3}\) rad = \(\frac{5\pi }{3}\cdot \frac{180\text{^{\circ}}}{\pi }=300\text{^{\circ}}\)
Express \(210\text{^{\circ}}\) using radians. Express \(11\pi \text{/}6\) rad using degrees.

જવાબ બતાવો

\(7\pi \text{/}6;\) 330°
Evaluate each of the following expressions.
1. \(\text{sin}(\frac{2\pi }{3})\)
2. \(\text{cos}(-\frac{5\pi }{6})\)
3. \(\text{tan}(\frac{15\pi }{4})\)
જવાબ બતાવો
1. On the unit circle, the angle \(\theta =\frac{2\pi }{3}\) corresponds to the point \((-\frac{1}{2},\frac{\sqrt{3}}{2}).\) Therefore, \(\text{sin}(\frac{2\pi }{3})=y=\frac{\sqrt{3}}{2}.\)
2. An angle \(\theta =-\frac{5\pi }{6}\) corresponds to a revolution in the negative direction, as shown. Therefore, \(\text{cos}(-\frac{5\pi }{6})=x=-\frac{\sqrt{3}}{2}.\)
3. An angle \(\theta =\frac{15\pi }{4}=2\pi +\frac{7\pi }{4}.\) Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of \(\frac{7\pi }{4}\) corresponds to the point \((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}),\) we can conclude that \(\text{tan}(\frac{15\pi }{4})=\frac{y}{x}=-1.\)
Evaluate \(\text{cos}(3\pi \text{/}4)\) and \(\text{sin}(\text{-}\pi \text{/}6).\)

જવાબ બતાવો

\(\text{cos}(3\pi \text{/}4)=\text{-}\sqrt{2}\text{/}2;\ \text{sin}(\text{-}\pi \text{/}6)=-1\text{/}2\)
A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is \(4\) ft from the ground and the angle between the ground and the ramp is to be \(10\text{^{\circ}},\) how long does the ramp need to be?

જવાબ બતાવો

Let \(x\) denote the length of the ramp. In the following image, we see that \(x\) needs to satisfy the equation \(\text{sin}(10\text{^{\circ}})=4\text{/}x.\) Solving this equation for \(x,\) we see that \(x=4\text{/}\text{sin}(10\text{^{\circ}})\approx 23.035\) ft.
A house painter wants to lean a \(20\)-ft ladder against a house. If the angle between the base of the ladder and the ground is to be \(60\text{^{\circ}},\) how far from the house should she place the base of the ladder?

જવાબ બતાવો

\(10\) ft
For each of the following equations, use a trigonometric identity to find all solutions.
1. \(1+\text{cos}(2\theta )=\text{cos}\ \theta\)
2. \(\text{sin}(2\theta )=\text{tan}\ \theta\)
જવાબ બતાવો
1. Using the double-angle formula for \(\text{cos}(2\theta ),\) we see that \(\theta\) is a solution of \[1+\text{cos}(2\theta )=\text{cos}\ \theta\]
  if and only if \[1+2{\text{cos}}^{2}\theta -1=\text{cos}\ \theta ,\]
  which is true if and only if \[2{\text{cos}}^{2}\theta -\text{cos}\ \theta =0.\]
  To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by \(\text{cos}\ \theta .\) The problem with dividing by \(\text{cos}\ \theta\) is that it is possible that \(\text{cos}\ \theta\) is zero. In fact, if we did divide both sides of the equation by \(\text{cos}\ \theta ,\) we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that \(\theta\) is a solution of this equation if and only if \[\text{cos}\ \theta (2\ \text{cos}\ \theta -1)=0.\]
  Since \(\text{cos}\ \theta =0\) when \[\theta =\frac{\pi }{2},\frac{\pi }{2}\pm \pi ,\frac{\pi }{2}\pm 2\pi \text{,\ldots ,}\]
  and \(\text{cos}\ \theta =1\text{/}2\) when \[\theta =\frac{\pi }{3},\frac{\pi }{3}\pm 2\pi \text{,\ldots }\ \text{or}\ \theta =-\frac{\pi }{3},-\frac{\pi }{3}\pm 2\pi \text{,\ldots ,}\]
  we conclude that the set of solutions to this equation is
  \[\theta =\frac{\pi }{2}+n\pi ,\theta =\frac{\pi }{3}+2n\pi ,\text{and}\ \theta =-\frac{\pi }{3}+2n\pi ,n=0,\pm 1,\pm 2,\text{\ldots }.\]
2. Using the double-angle formula for \(\text{sin}(2\theta )\) and the ratio identity for \(\text{tan}(\theta ),\) the equation can be written as \[2\ \text{sin}\ \theta \ \text{cos}\ \theta =\frac{\text{sin}\ \theta }{\text{cos}\ \theta }.\]
  To solve this equation, we multiply both sides by \(\text{cos}\ \theta\) to eliminate the denominator, and say that if \(\theta\) satisfies this equation, then \(\theta\) satisfies the equation \[2\ \text{sin}\ \theta {\text{cos}}^{2}\theta -\text{sin}\ \theta =0.\]
  However, we need to be a little careful here. Even if \(\theta\) satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by \(\text{cos}\ \theta .\) However, if \(\text{cos}\ \theta =0,\) we cannot divide both sides of the equation by \(\text{cos}\ \theta .\) Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor \(\text{sin}\ \theta\) out of both terms on the left-hand side instead of dividing both sides of the equation by \(\text{sin}\ \theta .\) Factoring the left-hand side of the equation, we can rewrite this equation as \[\text{sin}\ \theta (2{\text{cos}}^{2}\theta -1)=0.\]
  Therefore, the solutions are given by the angles \(\theta\) such that \(\text{sin}\ \theta =0\) or \({\text{cos}}^{2}\theta =1\text{/}2.\) The solutions of the first equation are \(\theta =0,\pm \pi ,\pm 2\pi \text{,\ldots .}\) The solutions of the second equation are \(\theta =\pi \text{/}4,(\pi \text{/}4)\pm (\pi \text{/}2),(\pi \text{/}4)\pm \pi \text{,\ldots .}\) After checking for extraneous solutions, the set of solutions to the equation is
  \[\theta =n\pi \ \text{and}\ \theta =\frac{\pi }{4}+\frac{n\pi }{2},n=0,\pm 1,\pm 2,\text{\ldots }.\]
Find all solutions to the equation \(\text{cos}(2\theta )=\text{sin}\ \theta .\)

જવાબ બતાવો

\(\theta =\frac{3\pi }{2}+2n\pi ,\frac{\pi }{6}+2n\pi ,\frac{5\pi }{6}+2n\pi\) for \(n=0,\pm 1,\pm 2\text{,\ldots }\)
Prove the trigonometric identity \(1+{\text{tan}}^{2}\theta ={\text{sec}}^{2}\theta .\)

જવાબ બતાવો

We start with the identity
\[{\text{sin}}^{2}\theta +{\text{cos}}^{2}\theta =1.\]
Dividing both sides of this equation by \({\text{cos}}^{2}\theta ,\) we obtain
\[\frac{{\text{sin}}^{2}\theta }{{\text{cos}}^{2}\theta }+1=\frac{1}{{\text{cos}}^{2}\theta }.\]
Since \(\text{sin}\ \theta \text{/}\text{cos}\ \theta =\text{tan}\ \theta\) and \(1\text{/}\text{cos}\ \theta =\text{sec}\ \theta ,\) we conclude that
\[{\text{tan}}^{2}\theta +1={\text{sec}}^{2}\theta .\]
Prove the trigonometric identity \(1+{\text{cot}}^{2}\theta ={\text{csc}}^{2}\theta .\)

જવાબ બતાવો

\(\begin{array}{lll}1+\theta & = & 1+\frac{\theta }{\theta } \\ & = & \frac{\theta }{\theta }+\frac{\theta }{\theta } \\ & = & \frac{\theta +\theta }{\theta } \\ & = & \frac{1}{\theta } \\ & = & \theta \end{array}\)
Sketch a graph of \(f(x)=3\ \text{sin}(2(x-\frac{\pi }{4}))+1.\)

જવાબ બતાવો

This graph is a horizontal compression by a factor of 2, a phase shift to the right by π/4 units, followed by a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of \(f\) is \(\pi .\)
Describe the relationship between the graph of \(f(x)=3\ \text{sin}(4x)-5\) and the graph of \(y=\text{sin}(x).\)

જવાબ બતાવો

To graph \(f(x)=3\ \text{sin}(4x)-5,\) the graph of \(y=\text{sin}(x)\) needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function \(f\) will have a period of \(\pi \text{/}2\) and an amplitude of 3.
\(240\text{^{\circ}}\)

જવાબ બતાવો

\(\frac{4\pi }{3}\ \text{rad}\)
\(15\text{^{\circ}}\)
\(-60\text{^{\circ}}\)

જવાબ બતાવો

\(\frac{\text{-}\pi }{3}\)
\(-225\text{^{\circ}}\)
\(330\text{^{\circ}}\)

જવાબ બતાવો

\(\frac{11\pi }{6}\ \text{rad}\)
\(\frac{\pi }{2}\ \text{rad}\)
\(\frac{7\pi }{6}\ \text{rad}\)

જવાબ બતાવો

\(210\text{^{\circ}}\)
\(\frac{11\pi }{2}\ \text{rad}\)
\(-3\pi \ \text{rad}\)

જવાબ બતાવો

\(-540\text{^{\circ}}\)
\(\frac{5\pi }{12}\ \text{rad}\)
\(\text{cos}(\frac{4\pi }{3})\)

જવાબ બતાવો

\(-0.5\)
\(\text{tan}(\frac{19\pi }{4})\)
\(\text{sin}(-\frac{3\pi }{4})\)

જવાબ બતાવો

\(-\frac{\sqrt{2}}{2}\)
\(\text{sec}(\frac{\pi }{6})\)
\(\text{sin}(\frac{\pi }{12})\)

જવાબ બતાવો

\(\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\text{cos}(\frac{5\pi }{12})\)
\(a=21,c=29\)
\(a=85.3,b=125.5\)

જવાબ બતાવો

a. \(c=151.7\) b. \(\text{sin}\ A=0.5623,\text{cos}\ A=0.8273,\text{tan}\ A=0.6797,\text{csc}\ A=1.778,\text{sec}\ A=1.209,\text{cot}\ A=1.471\)
\(b=40,c=41\)
\(a=84,b=13\)

જવાબ બતાવો

a. \(c=85\) b. \(\text{sin}\ A=\frac{84}{85},\text{cos}\ A=\frac{13}{85},\text{tan}\ A=\frac{84}{13},\text{csc}\ A=\frac{85}{84},\text{sec}\ A=\frac{85}{13},\text{cot}\ A=\frac{13}{84}\)
\(b=28,c=35\)
\(P(\frac{7}{25},y),y>0\)

જવાબ બતાવો

a. \(y=\frac{24}{25}\) b. \(\text{sin}\ \theta =\frac{24}{25},\text{cos}\ \theta =\frac{7}{25},\text{tan}\ \theta =\frac{24}{7},\text{csc}\ \theta =\frac{25}{24},\text{sec}\ \theta =\frac{25}{7},\text{cot}\ \theta =\frac{7}{24}\)
\(P(\frac{-15}{17},y),y<0\)
\(P(x,\frac{\sqrt{7}}{3}),x<0\)

જવાબ બતાવો

a. \(x=\frac{\text{-}\sqrt{2}}{3}\) b. \(\text{sin}\ \theta =\frac{\sqrt{7}}{3},\text{cos}\ \theta =\frac{\text{-}\sqrt{2}}{3},\text{tan}\ \theta =\frac{\text{-}\sqrt{14}}{2},\text{csc}\ \theta =\frac{3\sqrt{7}}{7},\text{sec}\ \theta =\frac{-3\sqrt{2}}{2},\text{cot}\ \theta =\frac{\text{-}\sqrt{14}}{7}\)
\(P(x,\frac{\text{-}\sqrt{15}}{4}),x>0\)
\({\text{tan}}^{2}x+\text{sin}\ x\ \text{csc}\ x\)

જવાબ બતાવો

\({\text{sec}}^{2}x\)
\(\text{sec}\ x\ \text{sin}\ x\ \text{cot}\ x\)
\(\frac{{\text{tan}}^{2}x}{{\text{sec}}^{2}x}\)

જવાબ બતાવો

\({\text{sin}}^{2}x\)

Symbols used here

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

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

\pi

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

\theta

theta
The usual name for an angle.

^\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.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

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

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Trigonometric Functions

Convert angle measures between degrees and radians.
Recognize the triangular and circular definitions of the basic trigonometric functions.
Write the basic trigonometric identities.
Identify the graphs and periods of the trigonometric functions.
Describe the shift of a sine or cosine graph from the equation of the function.
Express
Express
On the unit circle, the angle

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

તમારા પોતાના પ્રયત્ન કરો

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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