Question

If \[\sin x = \dfrac{1}{3}\] , how do you find $\cos 2x$ ?

Answer 1

Hint: As to solve this question. We should know about trigonometric identity.
Trigonometric identity: It is equalities that involve trigonometric function and true for every value of the occurring variable for which both sides of the equality are defined.
Some trigonometric identity:
 ${\sin ^2}\theta + {\cos ^2}\theta = 1$
 $\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)$

Complete step by step solution:
As given \[\sin x = \dfrac{1}{3}\] . we have to find $\cos 2x$ .
As we know that,
 $\Rightarrow \cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)$
and
 $\Rightarrow 1 = {\sin ^2}(x) + {\cos ^2}(x)$
We can write it as,
 $\Rightarrow {\cos ^2}(x) = 1 - {\sin ^2}(x)$ ………………. $(1)$
So:
 $\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)$
Keeping values form $(1)$ in it.
 $\Rightarrow 1 - {\sin ^2}(x) - {\sin ^2}(x)$
 $\Rightarrow 1 - 2{\sin ^2}(x)$
Keeping values as given in this question:
 $\Rightarrow \cos (2x) = 1 - {\sin ^2}(x)$
 $\Rightarrow 1 - 2{\left( {\dfrac{1}{3}} \right)^2}$
 $\Rightarrow 1 - \dfrac{2}{9} = \dfrac{7}{9}$

Hence, we had calculated $\cos (2x) = \dfrac{7}{9}$ .

Note: The inverse functions are partial inverse functions for the trigonometric function. For example the inverse for the sine, know as the inverse sine $({\sin ^{ - 1}})$ or arcsine (arcsin or asin), satisfy;
 $\sin (\arcsin x) = x$ for $\left| x \right| \leqslant 1$
 $\sin (\arcsin x) = x$ for $\left| x \right| \leqslant \dfrac{\pi }{2}$