Prove the following-cos 9 x -cos 5 x -sin 17 x -sin 3 x -sin 2 x -cos 10 x

We have L.H.S. = cos 9x cos 5x/sin 17x sin 3x = 2 sin ( 9x+5x/2)sin(9x 5x/2)/2 cos (17x+3x/2) sin (17x 3x/2) [ ∵cos C cos D = 2 sinC+D/2sinC D/2; s ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

Prove the fol...

Question

Prove the following:
$\frac{cos 9x - cos 5x}{sin 17x - sin 3x} = - \frac{sin 2x}{cos 10x}$

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Solution

We have

L.H.S. = $\frac{cos 9x - cos 5x}{sin 17x - sin 3x}$

= $\frac{- 2 sin (\frac{9 x + 5 x}{2}) sin (\frac{9 x - 5 x}{2})}{2 cos (\frac{17 x + 3 x}{2}) sin (\frac{17 x - 3 x}{2})}$

$⎡ ⎣ \begin{matrix} ∵ cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2} sin C- sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2} \end{matrix} ⎤ ⎦$

= $\frac{-2 sin 7x sin 2x}{2 cos 10 x sin 7x} = - \frac{sin 2x}{cos 10 x}$

= R.H.S.

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Prove the following: cos 9x−cos 5xsin 17x−sin 3x=−sin 2xcos 10x

We have L.H.S. = cos 9x−cos 5xsin 17x−sin 3x = −2sin(9x+5x2)sin(9x−5x2)2cos(17x+3x2)sin(17x−3x2) ⎡⎣∵cos C - cos D=−2sinC+D2sinC−D2sin C- sin D=2cosC+D2sinC−D2⎤⎦ = -2 sin 7x sin 2x2 cos 10 x sin 7x=−sin 2xcos 10 x = R.H.S.

Prove the following:
$\frac{cos 9x - cos 5x}{sin 17x - sin 3x} = - \frac{sin 2x}{cos 10x}$