Pre-Calculus - Sum and Difference Identities For Cosine

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

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

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

$\frac{-1}{2}$

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

$\frac{1}{2}$

Explanation

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

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

Yes it is an identity.

No it is not an identity.

It cannot be determined from the given information.

Explanation

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

and due to this inequality, this is not an identity

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

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

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

$-\frac{1}{2}$

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

$\frac{1}{2}$

Explanation

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

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

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

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

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

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

$\frac{1}{2}$

Explanation

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