Trigonometry Problem- 4th Dec'15

Question

If !!2Sin^2(pi/2Cos^2x) = 1 - Cos(piSin2x)!!

Then !!Cos2x = ?!!

!!a)" "1/5!!

!!b)" "3/5!!

!!c)" "4/5!!

!!d)" "!!None

Solution

In the given equation:

!!2Sin^2(pi/2Cos^2x) = 1 - Cos(piSin2x)!!

!!=> 2Sin^2(pi/2Cos^2x) = 2Sin^2((pi/2)Sin2x)!!

Hence, we can equate

!!(pi/2)Cos^2x = (pi/2)Sin2x!!

!!=> Cos^2x = Sin2x!!

!!=> Cos^2x = 2SinxCosx!!

This will give !!tanx = 1/2!!

Now we know that, !!Cos2x = (1 - tan^2x)/(1 + tan^2x)!!

!!Cos2x = (1 - 1/4)/(1 + 1/4) = 3/5!!

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