# express cos (x - y)

• Apr 17th 2006, 07:27 AM
ling_c_0202
express cos (x - y)
Could anyone give me some hints? I can't think the first step~

Given that $\displaystyle \sin x + \ sin y = a$,

$\displaystyle \cos x + \ cos y = b$,

express the following in therms of a and b:

(a) cos (x - y)

(b) sin (x - y)
• Apr 17th 2006, 07:50 AM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202
Could anyone give me some hints? I can't think the first step~

Given that $\displaystyle \sin x + \ sin y = a$,

$\displaystyle \cos x + \ cos y = b$,

express the following in therms of a and b:

(a) cos (x - y)

(b) sin (x - y)

(a)
$\displaystyle a=\sin x+\sin y$
Thus,
$\displaystyle a^2=\sin^2x+\sin^2y+2\sin x\sin y$
Thus,
$\displaystyle b^2=\cos^2x+\cos^2y+2\cos x\cos y$
Thus,
$\displaystyle a^2+b^2=\sin^2x+\cos^2x+\sin^2y+$$\displaystyle \cos^2y+2\sin x\sin y+2\cos x\cos y$
Thus,
$\displaystyle a^2+b^2=1+1+2(\cos x\cos y+\sin x\sin y)$
Thus,
$\displaystyle \frac{a^2+b^2}{2}-1=\cos(x-y)$
• Apr 17th 2006, 07:58 AM
rgep
Try multiplying each relation by $\displaystyle \sqrt{\frac12}$ and adding and subtracting.
• Apr 18th 2006, 08:16 AM
ling_c_0202
Thank you very much!! ^^