Example Question 9

Example corresponding to question 9 in trigonometry_2.

Example 1


What is $\sin x$ if $2 \tan x + 2 = 5$ and $0 \lt x \lt \dfrac{\pi}{2}$
  1. $\dfrac{2}{\sqrt {13}}$
  2. $\dfrac{3}{\sqrt {13}}$
  3. $-\dfrac{3}{\sqrt {13}}$
  4. $\dfrac{2}{3}$
  5. $\dfrac{4}{\sqrt {13}}$

Solution


  1. We first use the given equation to find $\tan \theta$.

    $2 \tan x + 2 = 5$

    $ 2\tan x = 5-2$

    $ \tan x = \dfrac{3}{2} $

  2. We first start with the identity $\tan x = \dfrac{\sin x}{\cos x}$ and substitute $\tan x$ by $\dfrac{3}{2}$ and write the equation

    $\dfrac{3}{2}=\dfrac{\sin x}{\cos x}$

  3. Square both sides

    $(\dfrac{3}{2})^2=(\dfrac{\sin x}{\cos x})^2$

    $\dfrac{9}{4}=\dfrac{\sin^2 x}{\cos^2 x}$

  4. Use the identity $\sin^2 x + \sin^2 x =1$ to substitute $\cos^2 x$ by $1-\sin^2 x$

    $\dfrac{9}{4}=\dfrac{\sin^2 x}{1-\sin^2 x}$

  5. Use cross product to rewrite the above equation as

    $9(1-\sin^2 x)=4\sin^2 x$ which gives $9-9\sin^2 x=4\sin^2 x$ and $13\sin^2 x=9$

    $\sin^2 x = \dfrac{9}{13}$

    $\sin x =\pm \sqrt{\dfrac{9}{13}}=\pm \dfrac{3}{\sqrt {13}}$

  6. The above equation has two solutions but $\sin x$ in quadrant $I$ is positive and therefore the positive solution is selected.

    $\sin x= \dfrac{3}{\sqrt {13}}$

    Answer B