Prove that- cos 9x - cos5xsin 17x-sin3x-sin 2xcos 10x

L.H.S=cos 9x cos 5xsin 17x sin 3x cos A cos B= 2 sin(A+B2) sin(A B2) sin A sin B=2 cos(A+B2) sin(A B2) ⇒ L.H.S= 2 sin(9x+5x2) sin(9x 5x2)2 cos(17x+3x2) si ...

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Standard XII

Mathematics

Circular Measurement of Angle

Prove that: c...

Question

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

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Solution

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

$cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2})$
$sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$
$\Rightarrow L . H . S = \frac{- 2 sin (\frac{9 x + 5 x}{2}) sin (\frac{9 x - 5 x}{2})}{2 cos (\frac{17 x + 3 x}{2}) sin (\frac{17 x - 3 x}{2})}$
$= \frac{- 2 sin (\frac{14 x}{2}) sin (\frac{4 x}{2})}{2 cos (\frac{20 x}{2}) sin (\frac{14 x}{2})}$
$= - \frac{2 sin (7 x) sin (2 x)}{2 cos (10 x) sin (7 x)} = - \frac{sin 2 x}{cos 10 x} = R . H . S .$

Hence proved.

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Circular Measurement of Angle

Standard XII Mathematics

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Prove that: cos9x−cos5xsin17x−sin3x=−sin2xcos10x

L.H.S=cos9x−cos5xsin17x−sin3x cosA−cosB=−2sin(A+B2)sin(A−B2) sinA−sinB=2cos(A+B2)sin(A−B2) ⇒L.H.S=−2sin(9x+5x2)sin(9x−5x2)2cos(17x+3x2)sin(17x−3x2) =−2sin(14x2)sin(4x2)2cos(20x2)sin(14x2) =−2sin(7x)sin(2x)2cos(10x)sin(7x)=−sin2xcos10x=R.H.S. Hence proved.

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