Prove that $$cos( \dfrac{3\pi}{4}+x)-cos(\dfrac{3\pi}{4}-x)=-\sqrt{2}sinx$$

Asked by Pragya Singh | 1 year ago |  62

##### Solution :-

We know that ,

cosA-cosB= -2sin $$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

$$cos( \dfrac{3\pi}{4}+x)-cos(\dfrac{3\pi}{4}-x)=-\sqrt{2}sinx$$

$$L.H.S= cos( \dfrac{3\pi}{4}+x)-cos(\dfrac{3\pi}{4}-x)$$

$$-2sin\{\dfrac{( \dfrac{3\pi}{4}+x)+(\dfrac{3\pi}{4}-x)}{2}\}$$

$$sin\{\dfrac{( \dfrac{3\pi}{4}+x)+(\dfrac{3\pi}{4}-x)}{2}\}$$

$$-2sin(\dfrac{3\pi}{4})sinx$$

Since sinx repeats the same value after an interval 2π ,

we have, $$sin(\dfrac{3\pi}{4})=sin(\pi-\dfrac{\pi}{4})$$

therefore,

L.H.S= -2sin$$\dfrac{\pi}{4}sinx$$

$$-2\times \dfrac{1}{\sqrt{2}}\times sinx$$

$$=-\sqrt{2}sinx$$

= R.H.S.

Hence proved.

Answered by Abhisek | 1 year ago

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