Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions

Tababar (40)

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1. Graph both sides of the identity \(\text{cot}\ \theta =\frac{1}{\tan \ \theta }.\) In other words, on the graphing calculator, graph \(y=\text{cot}\ \theta\) and \(y=\frac{1}{\tan \ \theta }.\)

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2. Verify \(\tan \ \theta \cos \ \theta =\sin \ \theta .\)

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   We will start on the left side, as it is the more complicated side:

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

3. Verify the identity \(\text{csc}\ \theta \ \cos \ \theta \ \tan \ \theta =1.\)

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   \[\begin{array}{lll}\text{csc}\ \theta \cos \ \theta \tan \ \theta & = & (\frac{1}{\sin \ \theta })\cos \ \theta (\frac{\sin \ \theta }{\cos \ \theta }) \\ & = & \frac{\cos \ \theta }{\sin \ \theta }(\frac{\sin \ \theta }{\cos \ \theta }) \\ & = & \frac{\sin \ \theta \cos \ \theta }{\sin \ \theta \cos \ \theta } \\ & = & 1\end{array}\]

4. Verify the following equivalency using the even-odd identities:

   \[(1+\sin \ x)[1+\sin (-x)]={\cos }^{2}x\]
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   Working on the left side of the equation, we have

   \[\begin{array}{llll}(1+\sin \ x)[1+\sin (-x)] & = & (1+\sin \ x)(1-\sin \ x) & \ \text{Since sin(-}x\text{)}=-\sin \ x \\ & = & 1-{\sin }^{2}x & \ \text{Difference of squares} \\ & = & {\cos }^{2}x & {\ \text{cos}}^{2}x=1-{\sin }^{2}x\end{array}\]

5. Verify the identity \(\frac{{\text{sec}}^{2}\theta -1}{{\text{sec}}^{2}\theta }={\sin }^{2}\theta\)

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   As the left side is more complicated, let’s begin there.

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

   There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

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

6. Show that \(\frac{\text{cot}\ \theta }{\text{csc}\ \theta }=\cos \ \theta .\)

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   \[\begin{array}{lll}\frac{\text{cot}\ \theta }{\text{csc}\ \theta } & = & \frac{\frac{\cos \ \theta }{\sin \ \theta }}{\frac{1}{\sin \ \theta }} \\ & = & \frac{\cos \ \theta }{\sin \ \theta }⋅\frac{\sin \ \theta }{1} \\ & = & \cos \ \theta \end{array}\]

7. Create an identity for the expression \(2\ \tan \ \theta \ \text{sec}\ \theta\) by rewriting strictly in terms of sine.

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   There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

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

   Thus,

   \[2\ \tan \ \theta \ \text{sec}\ \theta =\frac{2\ \sin \ \theta }{1-{\sin }^{2}\ \theta }\]

8. Verify the identity:

   \[\frac{{\sin }^{2}(-\theta )-{\cos }^{2}(-\theta )}{\sin (-\theta )-\cos (-\theta )}=\cos \ \theta -\sin \ \theta\]
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   Let’s start with the left side and simplify:

   \[\begin{array}{llll}\frac{{\sin }^{2}(-\theta )-{\cos }^{2}(-\theta )}{\sin (-\theta )-\cos (-\theta )} & = & \frac{{[\sin (-\theta )]}^{2}-{[\cos (-\theta )]}^{2}}{\sin (-\theta )-\cos (-\theta )} & \\ & = & \frac{{(-\sin \ \theta )}^{2}-{(\cos \ \theta )}^{2}}{-\sin \ \theta -\cos \ \theta } & \ \sin (-x)=-\sin \ x\ \text{and}\ \cos (-x)=\cos \ x \\ & = & \frac{{(\sin \ \theta )}^{2}-{(\cos \ \theta )}^{2}}{-\sin \ \theta -\cos \ \theta } & \ \text{Difference of squares} \\ & = & \frac{(\sin \ \theta -\cos \ \theta )(\sin \ \theta +\cos \ \theta )}{-(\sin \ \theta +\cos \ \theta )} & \\ & = & \frac{(\sin \ \theta -\cos \ \theta )(\sin \ \theta +\cos \ \theta )}{-(\sin \ \theta +\cos \ \theta )} & \\ & = & \cos \ \theta -\sin \ \theta & \end{array}\]

9. Verify the identity \(\frac{{\sin }^{2}\theta -1}{\tan \ \theta \ \sin \ \theta -\tan \ \theta }=\frac{\sin \ \theta +1}{\tan \ \theta }.\)

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   \(\begin{array}{lll}\frac{{\sin }^{2}\theta -1}{\tan \ \theta \sin \ \theta -\tan \ \theta } & = & \frac{(\sin \ \theta +1)(\sin \ \theta -1)}{\tan \ \theta (\sin \ \theta -1)} \\ & = & \frac{\sin \ \theta +1}{\tan \ \theta }\end{array}\)

10. Verify the identity: \((1-{\cos }^{2}x)(1+{\text{cot}}^{2}x)=1.\)

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    We will work on the left side of the equation.

    \[\begin{array}{llll}(1-{\cos }^{2}x)(1+{\text{cot}}^{2}x) & = & (1-{\cos }^{2}x)(1+\frac{{\cos }^{2}x}{{\sin }^{2}x}) & \\ & = & (1-{\cos }^{2}x)(\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}) & \ \text{Find the common denominator}. \\ & = & (1-{\cos }^{2}x)(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}) & \\ & = & ({\sin }^{2}x)(\frac{1}{{\sin }^{2}x}) & \\ & = & 1 & \end{array}\]

11. Write the following trigonometric expression as an algebraic expression: \(2{\cos }^{2}\theta +\cos \ \theta -1.\)

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    Notice that the pattern displayed has the same form as a standard quadratic expression, \(a{x}^{2}+bx+c.\) Letting \(\cos \ \theta =x,\) we can rewrite the expression as follows:

    \[2{x}^{2}+x-1\]

    This expression can be factored as \((2x-1)(x+1).\) If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for \(x.\) At this point, we would replace \(x\) with \(\cos \ \theta\) and solve for \(\theta .\)

12. Rewrite the trigonometric expression using the difference of squares: \(4\ {\cos }^{2}\theta -1.\)

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    Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares.

    \[\begin{array}{lll}4\ {\cos }^{2}\theta -1 & = & {(2\ \cos \ \theta )}^{2}-1 \\ & = & (2\ \cos \ \theta -1)(2\ \cos \ \theta +1)\end{array}\]

13. Rewrite the trigonometric expression using the difference of squares: \(25-9\ {\sin }^{2}\ \theta .\)

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    This is a difference of squares formula: \(25-9\ {\sin }^{2}\ \theta =(5-3\ \sin \ \theta )(5+3\ \sin \ \theta ).\)

14. Simplify the expression by rewriting and using identities:

    \[{\text{csc}}^{2}\theta -{\text{cot}}^{2}\theta\]
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    We can start with the Pythagorean identity.

    \[1+{\text{cot}}^{2}\theta ={\text{csc}}^{2}\theta\]

    Now we can simplify by substituting \(1+{\text{cot}}^{2}\theta\) for \({\text{csc}}^{2}\theta .\) We have

    \[\begin{array}{lll}{\text{csc}}^{2}\theta -{\text{cot}}^{2}\theta & = & 1+{\text{cot}}^{2}\theta -{\text{cot}}^{2}\theta \\ & = & 1\end{array}\]

15. Use algebraic techniques to verify the identity: \(\frac{\cos \ \theta }{1+\sin \ \theta }=\frac{1-\sin \ \theta }{\cos \ \theta }.\)

    (Hint: Multiply the numerator and denominator on the left side by \(1-\sin \ \theta .)\)

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

16. We know \(g(x)=\cos \ x\) is an even function, and \(f(x)=\sin \ x\) and \(h(x)=\tan \ x\) are odd functions. What about \(G(x)={\cos }^{2}x,F(x)={\sin }^{2}x,\) and \(H(x)={\tan }^{2}x?\) Are they even, odd, or neither? Why?

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    All three functions, \(F\),\(G\),\(\) and \(H\),\(\) are even.

    This is because \(F(-x)=\sin (-x)\sin (-x)=(-\sin \ x)(-\sin \ x)={\sin }^{2}x=F(x)\),\(G(-x)=\cos (-x)\cos (-x)=\cos \ x\cos \ x={\cos }^{2}x=G(x)\) and \(H(-x)=\tan (-x)\tan (-x)=(-\tan \ x)(-\tan \ x)={\tan }^{2}x=H(x).\)

17. Examine the graph of \(f(x)=\text{sec}\ x\) on the interval \([-\pi ,\pi ].\) How can we tell whether the function is even or odd by only observing the graph of \(f(x)=\text{sec}\ x?\)

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18. After examining the reciprocal identity for \(\text{sec}\ t,\) explain why the function is undefined at certain points.

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    When \(\cos \ t=0,\) then \(\text{sec}\ t=\frac{1}{0},\) which is undefined.

19. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from \({\sin }^{2}t+{\cos }^{2}t=1\) to the other forms.

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20. \(\sin \ x\ \cos \ x\ \text{sec}\ x\)

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    \(\sin \ x\)

21. \(\sin (-x)\ \cos (-x)\ \text{csc}(-x)\)

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22. \(\tan \ x\ \sin \ x+\text{sec}\ x\ {\cos }^{2}x\)

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    \(\text{sec}\ x\)

23. \(\text{csc}\ x+\cos \ x\ \text{cot}(-x)\)

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24. \(\frac{\text{cot}\ t+\tan \ t}{\text{sec}(-t)}\)

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    \(\text{csc}\ t\)

25. \(3\ {\sin }^{3}\ t\ \text{csc}\ t+{\cos }^{2}\ t+2\ \cos (-t)\cos \ t\)

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26. \(-\tan (-x)\text{cot}(-x)\)

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    \(-1\)

27. \(\frac{-\sin (-x)\cos \ x\ \text{sec}\ x\ \text{csc}\ x\ \tan \ x}{\text{cot}\ x}\)

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28. \(\frac{1+{\tan }^{2}\theta }{{\text{csc}}^{2}\theta }+{\sin }^{2}\theta +\frac{1}{{\text{sec}}^{2}\theta }\)

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

29. \((\frac{\tan \ x}{{\text{csc}}^{2}x}+\frac{\tan \ x}{{\text{sec}}^{2}x})(\frac{1+\tan \ x}{1+\text{cot}\ x})-\frac{1}{{\cos }^{2}x}\)

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30. \(\frac{1-{\cos }^{2}\ x}{{\tan }^{2}\ x}+2\ {\sin }^{2}\ x\)

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    \({\sin }^{2}x+1\)

31. \(\frac{\tan \ x+\text{cot}\ x}{\text{csc}\ x};\ \cos \ x\)

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32. \(\frac{\text{sec}\ x+\text{csc}\ x}{1+\tan \ x};\ \sin \ x\)

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    \(\frac{1}{\sin \ x}\)

33. \(\frac{\cos \ x}{1+\sin \ x}+\tan \ x;\ \cos \ x\)

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34. \(\frac{1}{\sin \ x\cos \ x}-\text{cot}\ x;\ \text{cot}\ x\)

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    \(\frac{1}{\text{cot}\ x}\)

35. \(\frac{1}{1-\cos \ x}-\frac{\cos \ x}{1+\cos \ x};\ \text{csc}\ x\)

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36. \((\text{sec}\ x+\text{csc}\ x)(\sin \ x+\cos \ x)-2-\text{cot}\ x;\ \tan \ x\)

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    \(\tan \ x\)

37. \(\frac{1}{\text{csc}\ x-\sin \ x};\ \text{sec}\ x\text{ and }\tan \ x\)

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38. \(\frac{1-\sin \ x}{1+\sin \ x}-\frac{1+\sin \ x}{1-\sin \ x};\ \text{sec}\ x\text{ and }\tan \ x\)

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    \(-4\text{sec}\ x\tan \ x\)

39. \(\tan \ x;\ \text{sec}\ x\)

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40. \(\text{sec}\ x;\ \text{cot}\ x\)

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    \(\pm \sqrt{\frac{1}{{\text{cot}}^{2}x}+1}\)

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.

In ka badan 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