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# Analytic Trig?

((cos x - cos y)/(sin x + sin y)) + ((sin x - sin y)/(cos x + cos y)).=0

I have no idea why though...Thanks.

### 3 Answers

- stephen mLv 41 decade agoFavorite Answer
Just put everything over a common denominator, and expand.

The numerator becomes:

(cos x - cos y)(cos x + cos y) + (sin x - sin y)(sin x + sin y)

= (cos x)^2 - (cos y)^2 + (sin x)^2 - (sin y)^2

= [(cos x)^2 + (sin x)^2] - [(cos y)^2 + (sin y)^2]

= 1 - 1 = 0.

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- 1 decade ago
To solve this, multiply the equation by both of the denominators. Then multiply out the equation. In the end you get cos^2 (x) + sin^2 (x) - (cos^2 (y) + sin^2 (y)) = 1 - 1 = 0

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- Scythian1950Lv 71 decade ago
Okay, multiply the first term by (cosx + cosy)/(cosx + cosy), and the 2nd term by (sinx + siny)/(sinx + siny), and adding the terms, we get:

(cosx)^2 - (cosy)^2 + (sinx)^2 - (siny)^2 divided by (sinx + siny)(cosx + cosy). Now, rerrange the top to get (cosx)^2 + (sinx)^2 - ( (cosy)^2 + (siny)^2 ), which reduces to 1 - 1, or 0. Recall that (sinx)^2 + (cosx)^2 = 1 as a very basic trig identity you should memorize.

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