Chapter EP, Problem 14.4.4EP

To determine

To find: the exact value of $\text{sin2θ}$ , $\text{cos2θ}$ , $\text{sin}\frac{\text{θ}}{2}$ and $\text{cos}\frac{\text{θ}}{2}$ of theexpression.

Answer to Problem 14.4.4EP

$\text{sin2θ}=0.71$ , $\text{cos2θ=0}\text{.704}$ , $\text{cos}\frac{\text{θ}}{2}=\frac{5\sqrt{26}}{26}$ and $\text{sin}\frac{\text{θ}}{\text{2}}=-0.196$ .

Explanation of Solution

Given:

$\text{sinθ=-}\frac{5}{13}$ . and ${\text{270}}^0{\text{<θ<360}}^{\text{0}}$ .

Concept used:

According to the trigonometric formula:

$\text{sin2θ=2sinθcosθ}$ and $\text{sinθ=2sin}\frac{\text{θ}}{\text{2}}\text{cos}\frac{\text{θ}}{\text{2}}$ .

${\text{cos2θ=2cos}}^{\text{2}}\text{θ-1}$ and ${\text{cosθ=2cos}}^{\text{2}}\frac{\text{θ}}{2}\text{-1}$ .

Calculation:

$\text{sinθ=-}\frac{5}{13}$ . and ${\text{270}}^0{\text{<θ<360}}^{\text{0}}$ .

  ${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$ .

  $\text{cosθ=±}\sqrt{\text{1-}{\left(\text{-}\frac{\text{5}}{\text{13}}\right)}^{\text{2}}}$ .

  $\text{cosθ=±}\sqrt{\text{1-}\frac{25}{169}}=\sqrt{\frac{144}{169}}=\pm \frac{12}{13}$ .

The value of $\text{cosθ}$ is positive since, the value of $\text{θ}$ lies in the fourth quadrant.

  $\text{cosθ=}\frac{12}{13}$ .

According to the trigonometric identities:

  $\text{sin2θ=2sinθcosθ}$ .

  $\Rightarrow \text{sin2θ=2}\times \left(\text{-}\frac{5}{13}\right)\times \frac{12}{13}=0.71$ .

  $\text{sin2θ}=0.71$ .

  ${\text{cos2θ=2cos}}^{\text{2}}\text{θ-1}$ ,

  $\Rightarrow \text{cos2θ=2}\left(\frac{144}{169}\right)\text{-1=0}\text{.704}$ .

  $\text{cos2θ=0}\text{.704}$ .

According to the trigonometric identities:

  $\frac{\frac{\text{12}}{\text{13}}\text{+1}}{\text{2}}{\text{=cos}}^{\text{2}}\frac{\text{θ}}{\text{2}}$ .

  $\frac{\text{25}}{26}{\text{=cos}}^{\text{2}}\frac{\text{θ}}{\text{2}}$ .

  $\text{cos}\frac{\text{θ}}{2}=\pm \sqrt{\frac{\text{25}}{26}}$ .

  $\text{cos}\frac{\text{θ}}{2}=\pm \frac{5\sqrt{26}}{26}$ .

The value of $\text{cos}\frac{\text{θ}}{2}$ is positive since, the value of $\text{θ}$ lies in the fourth quadrant.

  $\text{cos}\frac{\text{θ}}{2}=\frac{5\sqrt{26}}{26}$ .

According to the trigonometric identities:

  $\text{sinθ=2sin}\frac{\text{θ}}{\text{2}}\text{cos}\frac{\text{θ}}{\text{2}}$ .

  $\text{-}\frac{5}{13}\text{=2sin}\frac{\text{θ}}{\text{2}}\times \frac{5\sqrt{26}}{26}$ .

  $\text{sin}\frac{\text{θ}}{\text{2}}=-0.196$ .

Hence, $\text{sin2θ}=0.71$ , $\text{cos2θ=0}\text{.704}$ , $\text{cos}\frac{\text{θ}}{2}=\frac{5\sqrt{26}}{26}$ and $\text{sin}\frac{\text{θ}}{\text{2}}=-0.196$ .