Prove that sin 3 x-sin x-cos 2 x-2 sin x

LHS = sin 3x sinx/cos2x =2 cos ( 3x + x/2 ) sin ( 3x x/2 )/cos 2 x [ ∵ sin C sin D=2cos ( C + D/2 )sin (C D/2 ) ] =2 cos 2x sin x/cos 2x = 2 sin x ...

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Standard XII

Mathematics

Equations with Solutions at Boundary Values

Prove that si...

Question

Prove that $\frac{sin 3 x - sin x}{cos 2 x} = 2 sin x$

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Solution

LHS = $\frac{sin 3 x - sin x}{cos 2 x}$

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

$[∵ sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})]$

= $\frac{2 cos 2 x sin x}{cos 2 x} = 2 sin x$ = RHS.

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Prove that sin3x−sinxcos2x=2 sinx

LHS = sin3x−sinxcos2x =2 cos (3x+x2 ) sin (3x−x2)cos 2 x [∵ sin C−sin D=2cos(C + D2)sin(C - D2)] =2 cos 2x sinxcos 2x=2 sin x = RHS.

Prove that $\frac{sin 3 x - sin x}{cos 2 x} = 2 sin x$

LHS = $\frac{sin 3 x - sin x}{cos 2 x}$

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

$[∵ sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})]$

= $\frac{2 cos 2 x sin x}{cos 2 x} = 2 sin x$ = RHS.