Question

If cos2x = cos60° × cos30° + sin60° × sin30°, find sin2x

Answer 1

Answer:  The answer is $\dfrac{1}{2}.$

Step-by-step explanation:  Given that

$\cos 2x=\cos 60^\circ\times \cos 30^\circ+\sin 60^\circ\times \sin 30^\circ.$

To find the value of sin2x, first we will evaluate the right hand side of the above equality and then we will use the following formula from trigonometry.

$\sin A=\sqrt{1-\cos^2A}.$

Now,

$\cos 2x\\\\\cos 60^\circ\times \cos 30^\circ+\sin 60^\circ\times \sin 30^\circ\\\\=\dfrac{1}{2}\times \dfrac{\sqrt 3}{2}+\dfrac{\sqrt 3}{2}\times \dfrac{1}{2}\\\\\\=\dfrac{\sqrt 3}{4}+\dfrac{\sqrt 3}{4}\\\\\\=\dfrac{\sqrt 3}{2}.$

Therefore,

$\sin 2x=\sqrt{1-\left(\dfrac{\sqrt 3}{2}\right)^2}=\sqrt{1-\dfrac{3}{4}}=\sqrt {\dfrac{1}{4}}=\dfrac{1}{2}.$

Thus, the required value is $\dfrac{1}{2}.$

Answer 2

Answer:

Step-by-step explanation: