Sum and Difference Identities

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. Using the formula for the cosine of the difference of two angles, find the exact value of \(\cos (\frac{5\pi }{4}-\frac{\pi }{6}).\)

   Gosi nzaghachi

   Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.

   \[\begin{array}{lll}\cos (\alpha -\beta ) & = & \cos \ \alpha \ \cos \ \beta +\sin \ \alpha \ \sin \ \beta \\ \cos (\frac{5\pi }{4}-\frac{\pi }{6}) & = & \cos (\frac{5\pi }{4})\cos (\frac{\pi }{6})+\sin (\frac{5\pi }{4})\sin (\frac{\pi }{6}) \\ & = & (-\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2})(\frac{1}{2}) \\ & = & -\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & = & \frac{-\sqrt{6}-\sqrt{2}}{4}\end{array}\]

   Keep in mind that we can always check the answer using a graphing calculator in radian mode.

2. Find the exact value of \(\cos (\frac{\pi }{3}-\frac{\pi }{4}).\)

   Gosi nzaghachi

   \(\frac{\sqrt{2}+\sqrt{6}}{4}\)

3. Find the exact value of \(\cos (75^{\circ}).\)

   Gosi nzaghachi

   As \(75^{\circ}=45^{\circ}+30^{\circ},\) we can evaluate \(\cos (75^{\circ})\) as \(\cos (45^{\circ}+30^{\circ}).\)

   \[\begin{array}{lll}\cos (\alpha +\beta ) & = & \cos \ \alpha \ \cos \ \beta -\sin \ \alpha \ \sin \ \beta \\ \cos (45^{\circ}+30^{\circ}) & = & \cos (45^{\circ})\cos (30^{\circ})-\sin (45^{\circ})\sin (30^{\circ}) \\ & = & \frac{\sqrt{2}}{2}(\frac{\sqrt{3}}{2})-\frac{\sqrt{2}}{2}(\frac{1}{2}) \\ & = & \frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & = & \frac{\sqrt{6}-\sqrt{2}}{4}\end{array}\]

   Keep in mind that we can always check the answer using a graphing calculator in degree mode.

4. Find the exact value of \(\cos (105^{\circ}).\)

   Gosi nzaghachi

   \(\frac{\sqrt{2}-\sqrt{6}}{4}\)

5. Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b.

   1. ⓐ \(\sin (45^{\circ}-30^{\circ})\)
   2. ⓑ \(\sin (135^{\circ}-120^{\circ})\)
   Gosi nzaghachi

   1. ⓐ Let’s begin by writing the formula and substitute the given angles. \[\begin{array}{lll}\sin (\alpha -\beta ) & = & \sin \ \alpha \ \cos \ \beta -\cos \ \alpha \ \sin \ \beta \\ \sin (45^{\circ}-30^{\circ}) & = & \sin (45^{\circ})\cos (30^{\circ})-\cos (45^{\circ})\sin (30^{\circ})\end{array}\]

      Next, we need to find the values of the trigonometric expressions.

      \[\sin (45^{\circ})=\frac{\sqrt{2}}{2},\ \cos (30^{\circ})=\frac{\sqrt{3}}{2},\ \cos (45^{\circ})=\frac{\sqrt{2}}{2},\ \sin (30^{\circ})=\frac{1}{2}\]

      Now we can substitute these values into the equation and simplify.

      \[\begin{array}{lll}\sin (45^{\circ}-30^{\circ}) & = & \frac{\sqrt{2}}{2}(\frac{\sqrt{3}}{2})-\frac{\sqrt{2}}{2}(\frac{1}{2}) \\ & = & \frac{\sqrt{6}-\sqrt{2}}{4}\end{array}\]
   2. ⓑ Again, we write the formula and substitute the given angles. \[\begin{array}{lll}\sin (\alpha -\beta ) & = & \sin \ \alpha \ \cos \ \beta -\cos \ \alpha \ \sin \ \beta \\ \sin (135^{\circ}-120^{\circ}) & = & \sin (135^{\circ})\cos (120^{\circ})-\cos (135^{\circ})\sin (120^{\circ})\end{array}\]

      Next, we find the values of the trigonometric expressions.

      \[\sin (135^{\circ})=\frac{\sqrt{2}}{2},\cos (120^{\circ})=-\frac{1}{2},\cos (135^{\circ})=\frac{\sqrt{2}}{2},\sin (120^{\circ})=\frac{\sqrt{3}}{2}\]

      Now we can substitute these values into the equation and simplify.

      \[\begin{array}{lll}\sin (135^{\circ}-120^{\circ}) & = & \frac{\sqrt{2}}{2}(-\frac{1}{2})-(-\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) \\ & = & \frac{-\sqrt{2}+\sqrt{6}}{4} \\ & = & \frac{\sqrt{6}-\sqrt{2}}{4}\end{array}\]

6. Find the exact value of \(\sin ({\cos }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{3}{5}).\) Then check the answer with a graphing calculator.

   Gosi nzaghachi

   The pattern displayed in this problem is \(\sin (\alpha +\beta ).\) Let \(\alpha ={\cos }^{-1}\frac{1}{2}\) and \(\beta ={\sin }^{-1}\frac{3}{5}.\) Then we can write

   \[\begin{array}{lll}\cos \ \alpha & = & \frac{1}{2},0\le \alpha \le \pi \\ \sin \ \beta & = & \frac{3}{5},-\frac{\pi }{2}\le \beta \le \frac{\pi }{2}\end{array}\]

   We will use the Pythagorean identities to find \(\sin \ \alpha\) and \(\cos \ \beta .\)

   \[\begin{array}{lll}\sin \ \alpha & = & \sqrt{1-{\cos }^{2}\alpha } \\ & = & \sqrt{1-\frac{1}{4}} \\ & = & \sqrt{\frac{3}{4}} \\ & = & \frac{\sqrt{3}}{2} \\ \cos \ \beta & = & \sqrt{1-{\sin }^{2}\beta } \\ & = & \sqrt{1-\frac{9}{25}} \\ & = & \sqrt{\frac{16}{25}} \\ & = & \frac{4}{5}\end{array}\]

   Using the sum formula for sine,

   \[\begin{array}{lll}\sin ({\cos }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{3}{5}) & = & \sin (\alpha +\beta ) \\ & = & \sin \ \alpha \ \cos \ \beta +\cos \ \alpha \ \sin \ \beta \\ & = & \frac{\sqrt{3}}{2}⋅\frac{4}{5}+\frac{1}{2}⋅\frac{3}{5} \\ & = & \frac{4\sqrt{3}+3}{10}\end{array}\]

7. Find the exact value of \(\tan (\frac{\pi }{6}+\frac{\pi }{4}).\)

   Gosi nzaghachi

   Let’s first write the sum formula for tangent and then substitute the given angles into the formula.

   \[\begin{array}{lll}\tan (\alpha +\beta ) & = & \frac{\tan \ \alpha +\tan \ \beta }{1-\tan \ \alpha \ \tan \ \beta } \\ \tan (\frac{\pi }{6}+\frac{\pi }{4}) & = & \frac{\tan (\frac{\pi }{6})+\tan (\frac{\pi }{4})}{1-(\tan (\frac{\pi }{6}))(\tan (\frac{\pi }{4}))}\end{array}\]

   Next, we determine the individual function values within the formula:

   \[\tan (\frac{\pi }{6})=\frac{1}{\sqrt{3}},\tan (\frac{\pi }{4})=1\]

   So we have

   \[\begin{array}{lll}\tan (\frac{\pi }{6}+\frac{\pi }{4}) & = & \frac{\frac{1}{\sqrt{3}}+1}{1-(\frac{1}{\sqrt{3}})(1)} \\ & = & \frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} \\ & = & \frac{1+\sqrt{3}}{\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}-1}) \\ & = & \frac{\sqrt{3}+1}{\sqrt{3}-1}\end{array}\]

8. Find the exact value of \(\tan (\frac{2\pi }{3}+\frac{\pi }{4}).\)

   Gosi nzaghachi

   \(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

9. Given \(\sin \ \alpha =\frac{3}{5},0<\alpha <\frac{\pi }{2},\cos \ \beta =-\frac{5}{13},\pi <\beta <\frac{3\pi }{2},\) find

   1. ⓐ \(\sin (\alpha +\beta )\)
   2. ⓑ \(\cos (\alpha +\beta )\)
   3. ⓒ \(\tan (\alpha +\beta )\)
   4. ⓓ \(\tan (\alpha -\beta )\)
   Gosi nzaghachi

   We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

   • ⓐ To find \(\sin (\alpha +\beta ),\) we begin with \(\sin \ \alpha =\frac{3}{5}\) and \(0<\alpha <\frac{\pi }{2}.\) The side opposite \(\alpha\) has length 3, the hypotenuse has length 5, and \(\alpha\) is in the first quadrant. See . Using the Pythagorean Theorem, we can find the length of side \(a\text{:}\) \[\begin{array}{lll}{a}^{2}+{3}^{2} & = & {5}^{2} \\ {a}^{2} & = & 16 \\ a & = & 4\end{array}\]

     Since \(\cos \ \beta =-\frac{5}{13}\) and \(\pi <\beta <\frac{3\pi }{2},\) the side adjacent to \(\beta\) is \(-5,\) the hypotenuse is 13, and \(\beta\) is in the third quadrant. See . Again, using the Pythagorean Theorem, we have

     \[\begin{array}{lll}{(-5)}^{2}+{a}^{2} & = & {13}^{2} \\ 25+{a}^{2} & = & 169 \\ {a}^{2} & = & 144 \\ a & = & \pm 12\end{array}\]

     Since \(\beta\) is in the third quadrant, \(a=-12.\)

     The next step is finding the cosine of \(\alpha\) and the sine of \(\beta .\) The cosine of \(\alpha\) is the adjacent side over the hypotenuse. We can find it from the triangle in : \(\cos \ \alpha =\frac{4}{5}.\) We can also find the sine of \(\beta\) from the triangle in , as opposite side over the hypotenuse: \(\sin \ \beta =-\frac{12}{13}.\) Now we are ready to evaluate \(\sin (\alpha +\beta ).\)

     \[\begin{array}{lll}\sin (\alpha +\beta ) & = & \sin \ \alpha \ \cos \ \beta +\cos \ \alpha \ \sin \ \beta \\ & = & (\frac{3}{5})(-\frac{5}{13})+(\frac{4}{5})(-\frac{12}{13}) \\ & = & -\frac{15}{65}-\frac{48}{65} \\ & = & -\frac{63}{65}\end{array}\]
   • ⓑ We can find \(\cos (\alpha +\beta )\) in a similar manner. We substitute the values according to the formula. \[\begin{array}{lll}\cos (\alpha +\beta ) & = & \cos \ \alpha \ \cos \ \beta -\sin \ \alpha \ \sin \ \beta \\ & = & (\frac{4}{5})(-\frac{5}{13})-(\frac{3}{5})(-\frac{12}{13}) \\ & = & -\frac{20}{65}+\frac{36}{65} \\ & = & \frac{16}{65}\end{array}\]
   • ⓒ For \(\tan (\alpha +\beta ),\) if \(\sin \ \alpha =\frac{3}{5}\) and \(\cos \ \alpha =\frac{4}{5},\) then \[\tan \ \alpha =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\]

     If \(\sin \ \beta =-\frac{12}{13}\) and \(\cos \ \beta =-\frac{5}{13},\) then

     \[\tan \ \beta =\frac{\frac{-12}{13}}{\frac{-5}{13}}=\frac{12}{5}\]

     Then,

     \[\begin{array}{lll}\tan (\alpha +\beta ) & = & \frac{\tan \ \alpha +\tan \ \beta }{1-\tan \ \alpha \ \tan \ \beta } \\ & = & \frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}(\frac{12}{5})} \\ & = & \frac{\ \frac{63}{20}}{-\frac{16}{20}} \\ & = & -\frac{63}{16}\end{array}\]
   • ⓓ To find \(\tan (\alpha -\beta ),\) we have the values we need. We can substitute them in and evaluate. \[\begin{array}{lll}\tan (\alpha -\beta ) & = & \frac{\tan \ \alpha -\tan \ \beta }{1+\tan \ \alpha \ \tan \ \beta } \\ & = & \frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}(\frac{12}{5})} \\ & = & \frac{-\frac{33}{20}}{\frac{56}{20}} \\ & = & -\frac{33}{56}\end{array}\]

10. Write \(\tan \ \frac{\pi }{9}\) in terms of its cofunction.

    Gosi nzaghachi

    The cofunction of \(\tan \ \theta =\text{cot}(\frac{\pi }{2}-\theta ).\) Thus,

    \[\begin{array}{lll}\tan (\frac{\pi }{9}) & = & \text{cot}(\frac{\pi }{2}-\frac{\pi }{9}) \\ & = & \text{cot}(\frac{9\pi }{18}-\frac{2\pi }{18}) \\ & = & \text{cot}(\frac{7\pi }{18})\end{array}\]

11. Write \(\sin \ \frac{\pi }{7}\) in terms of its cofunction.

    Gosi nzaghachi

    \(\cos (\frac{5\pi }{14})\)

12. Verify the identity \(\sin (\alpha +\beta )+\sin (\alpha -\beta )=2\ \sin \ \alpha \ \cos \ \beta .\)

    Gosi nzaghachi

    We see that the left side of the equation includes the sines of the sum and the difference of angles.

    \[\begin{array}{lll}\sin (\alpha +\beta ) & = & \sin \ \alpha \ \cos \ \beta +\cos \ \alpha \ \sin \ \beta \\ \sin (\alpha -\beta ) & = & \sin \ \alpha \ \cos \ \beta -\cos \ \alpha \ \sin \ \beta \end{array}\]

    We can rewrite each using the sum and difference formulas.

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

    We see that the identity is verified.

13. Verify the following identity.

    \[\frac{\sin (\alpha -\beta )}{\cos \ \alpha \ \cos \ \beta }=\tan \ \alpha -\tan \ \beta\]
    Gosi nzaghachi

    We can begin by rewriting the numerator on the left side of the equation.

    \[\begin{array}{llll}\frac{\sin (\alpha -\beta )}{\cos \ \alpha \ \cos \ \beta } & = & \frac{\sin \ \alpha \ \cos \ \beta -\cos \ \alpha \sin \ \beta }{\cos \ \alpha \cos \ \beta } & \\ & = & \frac{\sin \ \alpha \ \cos \ \beta }{\cos \ \alpha \ \cos \ \beta }-\frac{\cos \ \alpha \ \sin \ \beta }{\cos \ \alpha \ \cos \ \beta } & \ \text{Rewrite using a common denominator}. \\ & = & \frac{\sin \ \alpha }{\cos \ \alpha }-\frac{\sin \ \beta }{\cos \ \beta } & \ \text{Cancel}. \\ & = & \tan \ \alpha -\tan \ \beta & \ \text{Rewrite in terms of tangent}.\end{array}\]

    We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.

14. Verify the identity: \(\tan (\pi -\theta )=-\tan \ \theta .\)

    Gosi nzaghachi

    \[\begin{array}{lll}\tan (\pi -\theta ) & = & \frac{\tan (\pi )-\tan \ \theta }{1+\tan (\pi )\tan \theta } \\ & = & \frac{0-\tan \ \theta }{1+0⋅\tan \ \theta } \\ & = & -\tan \ \theta \end{array}\]

15. Let \({L}_{1}\) and \({L}_{2}\) denote two non-vertical intersecting lines, and let \(\theta\) denote the acute angle between \({L}_{1}\) and \({L}_{2}.\) See . Show that

    \[\tan \ \theta =\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\]

    where \({m}_{1}\) and \({m}_{2}\) are the slopes of \({L}_{1}\) and \({L}_{2}\) respectively. (Hint: Use the fact that \(\tan \ {\theta }_{1}={m}_{1}\) and \(\tan \ {\theta }_{2}={m}_{2}.\) )

    Gosi nzaghachi

    Using the difference formula for tangent, this problem does not seem as daunting as it might.

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

16. For a climbing wall, a guy-wire \(R\) is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire \(S\) attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle \(\alpha\) between the wires. See .

    Gosi nzaghachi

    Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that \(\tan \ \beta =\frac{47}{50},\) and \(\tan (\beta -\alpha )=\frac{40}{50}=\frac{4}{5}.\) We can then use difference formula for tangent.

    \[\tan (\beta -\alpha )=\frac{\tan \ \beta -\tan \ \alpha }{1+\tan \ \beta \tan \ \alpha }\]

    Now, substituting the values we know into the formula, we have

    \[\begin{array}{lll}\frac{4}{5} & = & \frac{\frac{47}{50}-\tan \ \alpha }{1+\frac{47}{50}\tan \ \alpha } \\ 4(1+\frac{47}{50}\tan \ \alpha ) & = & 5(\frac{47}{50}-\tan \ \alpha )\end{array}\]

    Use the distributive property, and then simplify the functions.

    \[\begin{array}{lll}4(1)+4(\frac{47}{50})\tan \ \alpha & = & 5(\frac{47}{50})-5\ \tan \ \alpha \\ 4+3.76\ \tan \ \alpha & = & 4.7-5\ \tan \ \alpha \\ 5\ \tan \ \alpha +3.76\ \tan \ \alpha & = & 0.7 \\ 8.76\ \tan \ \alpha & = & 0.7 \\ \tan \ \alpha & \approx & 0.07991 \\ {\tan }^{-1}(0.07991) & \approx & .079741\end{array}\]

    Now we can calculate the angle in degrees.

    \[\alpha \approx 0.079741(\frac{180}{\pi })\approx 4.57^{\circ}\]

17. Explain the basis for the cofunction identities and when they apply.

    Gosi nzaghachi

    The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures \(x,\) the second angle measures \(\frac{\pi }{2}-x.\) Then \(\sin x=\cos (\frac{\pi }{2}-x).\) The same holds for the other cofunction identities. The key is that the angles are complementary.

18. Is there only one way to evaluate \(\cos (\frac{5\pi }{4})?\) Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

19. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for \(f(x)=\sin (x)\) and \(g(x)=\cos (x).\) (Hint: \(0-x=-x\) )

    Gosi nzaghachi

    \(\sin (-x)=-\sin x,\) so \(\sin x\) is odd. \(\cos (-x)=\cos (0-x)=\cos x,\) so \(\cos x\) is even.

20. \(\cos (\frac{7\pi }{12})\)

21. \(\cos (\frac{\pi }{12})\)

    Gosi nzaghachi

    \(\frac{\sqrt{2}+\sqrt{6}}{4}\)

22. \(\sin (\frac{5\pi }{12})\)

23. \(\sin (\frac{11\pi }{12})\)

    Gosi nzaghachi

    \(\frac{\sqrt{6}-\sqrt{2}}{4}\)

24. \(\tan (-\frac{\pi }{12})\)

25. \(\tan (\frac{19\pi }{12})\)

    Gosi nzaghachi

    \(-2-\sqrt{3}\)

26. \(\sin (x+\frac{11\pi }{6})\)

27. \(\sin (x-\frac{3\pi }{4})\)

    Gosi nzaghachi

    \(-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\)

28. \(\cos (x-\frac{5\pi }{6})\)

29. \(\cos (x+\frac{2\pi }{3})\)

    Gosi nzaghachi

    \(-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\)

30. \(\text{csc}(\frac{\pi }{2}-t)\)

31. \(\text{sec}(\frac{\pi }{2}-\theta )\)

    Gosi nzaghachi

    \(\text{csc}\theta\)

32. \(\text{cot}(\frac{\pi }{2}-x)\)

33. \(\tan (\frac{\pi }{2}-x)\)

    Gosi nzaghachi

    \(\text{cot}x\)

34. \(\sin (2x)\ \cos (5x)-\sin (5x)\ \cos (2x)\)

35. \(\frac{\tan (\frac{3}{2}x)-\tan (\frac{7}{5}x)}{1+\tan (\frac{3}{2}x)\tan (\frac{7}{5}x)}\)

    Gosi nzaghachi

    \(\tan (\frac{x}{10})\)

36. Given that \(\sin \ a=\frac{2}{3}\) and \(\cos \ b=-\frac{1}{4},\) with \(a\) and \(b\) both in the interval \([\frac{\pi }{2},\pi ),\) find \(\sin (a+b)\) and \(\cos (a-b).\)

37. Given that \(\sin \ a=\frac{4}{5},\) and \(\cos \ b=\frac{1}{3},\) with \(a\) and \(b\) both in the interval \([0,\frac{\pi }{2}),\) find \(\sin (a-b)\) and \(\cos (a+b).\)

    Gosi nzaghachi

    \(\begin{array}{lllll}\sin (a-b) & = & (\frac{4}{5})(\frac{1}{3})-(\frac{3}{5})(\frac{2\sqrt{2}}{3}) & = & \frac{4-6\sqrt{2}}{15} \\ \cos (a+b) & = & (\frac{3}{5})(\frac{1}{3})-(\frac{4}{5})(\frac{2\sqrt{2}}{3}) & = & \frac{3-8\sqrt{2}}{15}\end{array}\)

38. \(\sin ({\cos }^{-1}(0)-{\cos }^{-1}(\frac{1}{2}))\)

39. \(\cos ({\cos }^{-1}(\frac{\sqrt{2}}{2})+{\sin }^{-1}(\frac{\sqrt{3}}{2}))\)

    Gosi nzaghachi

    \(\frac{\sqrt{2}-\sqrt{6}}{4}\)

40. \(\tan ({\sin }^{-1}(\frac{1}{2})-{\cos }^{-1}(\frac{1}{2}))\)

Jiri gị onwe gị

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.

Oge Trigonometry

The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions