express cos (x - y)

**ling_c_0202** · April 17th 2006, 07:27 AM

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

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

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

express the following in therms of a and b:

(a) cos (x - y)

(b) sin (x - y)

**ThePerfectHacker** · April 17th 2006, 07:50 AM

Originally Posted by ling_c_0202

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

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

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

express the following in therms of a and b:

(a) cos (x - y)

(b) sin (x - y)

(a)
$a=\sin x+\sin y$
Thus,
$a^2=\sin^2x+\sin^2y+2\sin x\sin y$
Thus,
$b^2=\cos^2x+\cos^2y+2\cos x\cos y$
Thus,
$a^2+b^2=\sin^2x+\cos^2x+\sin^2y+$ $\cos^2y+2\sin x\sin y+2\cos x\cos y$
Thus,
$a^2+b^2=1+1+2(\cos x\cos y+\sin x\sin y)$
Thus,
$\frac{a^2+b^2}{2}-1=\cos(x-y)$

**rgep** · April 17th 2006, 07:58 AM

Try multiplying each relation by $\sqrt{\frac12}$ and adding and subtracting.

**ling_c_0202** · April 18th 2006, 08:16 AM

Thank you very much!! ^^

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