Proving Trig Identities

**magentarita** · August 26th 2008, 05:42 AM

Prove each trig identity below.

(1) cos(pi/2 + x) = -sin(x)

(2) cos(A + B) + cos(A - B) = 2 cosA cos B

**Moo** · August 26th 2008, 06:03 AM

Originally Posted by magentarita

Prove each trig identity below.

(1) cos(pi/2 + x) = -sin(x)

(2) cos(A + B) + cos(A - B) = 2 cosA cos B

(1) -> use this formula :
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ , and remember these two values :
$\cos \tfrac \pi 2=0 \text{ and } \sin \tfrac \pi 2=1$

(2) -> use the formula above and this formula :
$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ (this is because cosine is an even function and sine an odd function)
Then, add up $\cos(a-b)+\cos(a+b)$

**Chop Suey** · August 26th 2008, 06:03 AM

$\cos{(A \pm B)} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$

For both questions, prove the LHS. I want to see your attempt and people will point out any mistakes, if any.

**magentarita** · August 26th 2008, 08:53 PM

Originally Posted by Moo

(1) -> use this formula :
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ , and remember these two values :
$\cos \tfrac \pi 2=0 \text{ and } \sin \tfrac \pi 2=1$

(2) -> use the formula above and this formula :
$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ (this is because cosine is an even function and sine an odd function)
Then, add up $\cos(a-b)+\cos(a+b)$

Thanks for the tips.

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