Find Exact Vaue of cos(2x)

Example on how to find the exact value of trigonometric functions corresponding to question 4 in trigonometry_2.

Example 1


What is the exact value of $\cos (2x)$ if $\sin x = \dfrac{1}{3}$ and $x$ is in quadrant $I$?

  1. $\dfrac{4\sqrt 2}{3}$

  2. $\dfrac{7}{9}$

  3. $\dfrac{2}{3}$

  4. $\dfrac{2\sqrt 2}{3}$

  5. $\dfrac{8}{9}$

Solution


  1. We first use the identity

    $\cos(2x)=\cos^2 x - \sin^2 x$

  2. We now use the identity $\cos^2 x = 1 - \sin ^2 x$ to rewrite $\cos (2x)$ as follows

    $\cos(2x)=1 - \sin ^2 x - \sin^2 x$

  3. Group like terms

    $\cos(2x)=1 - 2 \sin ^2 x$

  4. Use $\sin x = \dfrac{1}{3}$ in the expression of $\cos (2x)$ to obtain

    $\cos(2x)=1 - 2(\dfrac{1}{3})^2 = 1 - 2 \dfrac{1}{9} = 1 - \dfrac{2}{9}$

    $= \dfrac{9-2}{9} = \dfrac{7}{9}$

    Answer B