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Precalculus : Sum and Difference Identities For Cosine

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Example Question #1 : Sum And Difference Identities For Cosine

Use the sum or difference identity to find the exact value: $cos255^\circ$ $\displaystyle cos255^\circ$

Possible Answers:

$\frac{-1}{2}$ $\displaystyle \frac{-1}{2}$

$\frac{\sqrt{2}+\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{2\sqrt{2}}{4}$ $\displaystyle \frac{2\sqrt{2}}{4}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

$\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$

Explanation:

Using the identity, we can break up the 255 into and then solve: $cos300\cdot cos45+sin300\cdot sin45\rightarrow \frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{-\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\rightarrow \frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle cos300\cdot cos45+sin300\cdot sin45\rightarrow \frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{-\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\rightarrow \frac{\sqrt{2}-\sqrt{6}}{4}$

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Example Question #1 : Trigonometric Identities

Is the following equation an identity? $cos(x-\pi)=cos(x)$ $\displaystyle cos(x-\pi)=cos(x)$

Possible Answers:

Yes it is an identity.

No it is not an identity.

It cannot be determined from the given information.

Correct answer:

No it is not an identity.

Explanation:

$cos(x-\pi)=cos(x)\rightarrow cosxcos\pi+sinxsin\pi=cosx\rightarrow cosx(-1)+sinx(0)=cosx\rightarrow -cosx\neq cosx$ $\displaystyle cos(x-\pi)=cos(x)\rightarrow cosxcos\pi+sinxsin\pi=cosx\rightarrow cosx(-1)+sinx(0)=cosx\rightarrow -cosx\neq cosx$

and due to this inequality, this is not an identity

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Example Question #1 : Sum And Difference Identities

Use the sum or difference identity to find the exact value: $\cos 255^{\circ}$ $\displaystyle \cos 255^{\circ}$

Possible Answers:

$\frac{\sqrt{2}+\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{2\sqrt{2}}{4}$ $\displaystyle \frac{2\sqrt{2}}{4}$

Correct answer:

$\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$

Explanation:

Using the identity, we can break up the 255 $\displaystyle 255$ into 300-45 $\displaystyle 300-45$ and then solve: $\cos 300\cdot \cos 45+\sin 300\cdot \sin 45\rightarrow \frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{-\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\rightarrow \frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \cos 300\cdot \cos 45+\sin 300\cdot \sin 45\rightarrow \frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{-\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\rightarrow \frac{\sqrt{2}-\sqrt{6}}{4}$ and so the correct answer is $\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$.

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Example Question #2 : Trigonometric Identities

Is the following equation an identity?

$\cos \left ( x-\pi \right )=\cos \left ( x \right )$ $\displaystyle \cos \left ( x-\pi \right )=\cos \left ( x \right )$

Possible Answers:

The answer cannot be determined from the information provided.

No it is not an identity

Yes it is an identity

Correct answer:

No it is not an identity

Explanation:

$\cos \left ( x-\pi \right )=\cos \left ( x \right )\rightarrow \cos x\cos \pi +\sin x\sin \pi =\cos x\rightarrow \cos x(-1)+\sin x(0)=\cos x\rightarrow -\cos x\neq \cos x$ $\displaystyle \cos \left ( x-\pi \right )=\cos \left ( x \right )\rightarrow \cos x\cos \pi +\sin x\sin \pi =\cos x\rightarrow \cos x(-1)+\sin x(0)=\cos x\rightarrow -\cos x\neq \cos x$

Due to this inequality, this is not an identity.

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Example Question #9 : Sum And Difference Identities

Use the sum or difference identity to find the exact value of $\sin 165^{\circ}$ $\displaystyle \sin 165^{\circ}$.

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{\sqrt{6}}4{}$ $\displaystyle \frac{\sqrt{6}}4{}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$ $\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$ $\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}$

$\frac{\sqrt{6}+\sqrt{2}}{4}$ $\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}$

Correct answer:

$\frac{\sqrt{6}-\sqrt{2}}{4}$ $\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}$

Explanation:

Here we break up the 165 $\displaystyle 165$ into 135+30 $\displaystyle 135+30$ and solve using the sin identity: $\sin 135\cdot \cos 30+\cos 135\cdot \sin 30\rightarrow \frac{\sqrt{2}}2{}\cdot \frac{\sqrt{3}}2{}+\frac{-\sqrt{2}}2{}\cdot \frac{1}{2}\rightarrow \frac{\sqrt{6}-\sqrt{2}}{4}$ $\displaystyle \sin 135\cdot \cos 30+\cos 135\cdot \sin 30\rightarrow \frac{\sqrt{2}}2{}\cdot \frac{\sqrt{3}}2{}+\frac{-\sqrt{2}}2{}\cdot \frac{1}{2}\rightarrow \frac{\sqrt{6}-\sqrt{2}}{4}$ and so here the credited answer is $\frac{\sqrt{6}-\sqrt{2}}{4}$ $\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}$.

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Precalculus : Sum and Difference Identities For Cosine

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Sum And Difference Identities For Cosine

Example Question #1 : Trigonometric Identities

Example Question #1 : Sum And Difference Identities

Example Question #2 : Trigonometric Identities

Example Question #9 : Sum And Difference Identities

All Precalculus Resources

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