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Doctors Medicines MCQs Q&A

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Use $sin x = cos (pi/2-x), cos x = sin ( pi/2 - x)$

$cos^(-1) cos a =a, sin^(-1) sin b = b$ and

$sin 2A - sin 2 B = 2 cos ( A + B ) sin ( A - B )$.

Let $u =cos^(-1) sinx + sin ^(-1)cos x$

$=cos^(-1) cos (pi/2-x) + sin ^(-1) sin (pi/2-x)$

$=(pi/2-x) + (pi/2-x)$

$=pi - 2x$

So, the given equation is

$A sin ( B x + C )$

$=cos (pi-2x)+sin(pi-2x)$

$=-cos 2x+ sin 2x$

$=sin 2x - sin (pi/2-2x)$

$= 2 cos (pi/4) sin ( 2x - pi/4)$

$=sqrt 2 sin ( 2x - pi/4)$

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