cos 2x =

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Both B and C

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Solution

The correct option is D
Both B and C

cos(A+B) = cosA cosB - sinA sinB

If A = B =x

cos(2x) = cosx.cosx - sinx.snx

= $c o s^{2} x - s i n^{2} x$

( $s i n^{2} x + c o s^{2} x$ ) = 1

Substituting $c o s^{2} x = 1 - s i n^{2} x$

= 1 - $s i n^{2} x - s i n^{2} x$

cos2x = $1 - 2 s i n^{2} x$

(substituting $s i n^{2} x = 1 - c o s^{2} x$ )

= $c o s^{2} x - (1 - c o s^{2} x)$

= cos2x = $2 c o s^{2} x - 1$

So, both the option B and C are correct.

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Similar questions

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Prove that

(i) $\frac{sin 2 x}{1 - cos 2 x} = cot x$

(ii) $\frac{1 - cos2x}{1 + cos 2 x} = {tan}^{2} x$

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