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Question: If \[\sin x = \dfrac{1}{3}\] , how do you find \(\cos 2x\) ?...

If sin⁡x=13\sin x = \dfrac{1}{3} , how do you find cos⁡2x\cos 2x ?

Explanation

Solution

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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
cos⁡(2x)=cos⁡2(x)−sin⁡2(x)\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)

Complete step by step solution:
As given sin⁡x=13\sin x = \dfrac{1}{3} . we have to find cos⁡2x\cos 2x .
As we know that,
⇒cos⁡(2x)=cos⁡2(x)−sin⁡2(x)\Rightarrow \cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)
and
⇒1=sin⁡2(x)+cos⁡2(x)\Rightarrow 1 = {\sin ^2}(x) + {\cos ^2}(x)
We can write it as,
⇒cos⁡2(x)=1−sin⁡2(x)\Rightarrow {\cos ^2}(x) = 1 - {\sin ^2}(x) ………………. (1)(1)
So:
cos⁡(2x)=cos⁡2(x)−sin⁡2(x)\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)
Keeping values form (1)(1) in it.
⇒1−sin⁡2(x)−sin⁡2(x)\Rightarrow 1 - {\sin ^2}(x) - {\sin ^2}(x)
⇒1−2sin⁡2(x)\Rightarrow 1 - 2{\sin ^2}(x)
Keeping values as given in this question:
⇒cos⁡(2x)=1−sin⁡2(x)\Rightarrow \cos (2x) = 1 - {\sin ^2}(x)
⇒1−2(13)2\Rightarrow 1 - 2{\left( {\dfrac{1}{3}} \right)^2}
⇒1−29=79\Rightarrow 1 - \dfrac{2}{9} = \dfrac{7}{9}

Hence, we had calculated cos⁡(2x)=79\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)({\sin ^{ - 1}}) or arcsine (arcsin or asin), satisfy;
sin⁡(arcsin⁡x)=x\sin (\arcsin x) = x for ∣x∣⩽1\left| x \right| \leqslant 1
sin⁡(arcsin⁡x)=x\sin (\arcsin x) = x for ∣x∣⩽π2\left| x \right| \leqslant \dfrac{\pi }{2}