Question

If $\sin \theta =\frac{1}{4}$and $\theta$is in second quadrant, find

(a) $\cos \theta$ (b) $\tan \theta$

(c) $\sin \left(2\theta \right)$ (d) $\cos \left(2\theta \right)$

(e)$\sin \left(\frac{1}{2}\theta \right)$

Short answer

(a) $\cos \theta =-\frac{\sqrt{15}}{4}$

(b) $\tan \theta =-\frac{1}{\sqrt{15}}$

(c)$\sin \left(2\theta \right)=-\frac{\sqrt{15}}{8}$

(d)$\cos \left(2\theta \right)=\frac{7}{8}$

(e) $\sin \left(\frac{1}{2}\theta \right)=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$

Step-by-step solution

Step 1. Given Information

Given $\sin \theta =\frac{1}{4}$and $\frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$.

Part (a) Step 1. Finding cosθ 

We know that

$$\begin{gathered} {\cos }^2\theta +{\sin }^2\theta =1 \\ {\cos }^2\theta +{\left(\frac{1}{4}\right)}^2=1 \\ {\cos }^2\theta =1-\frac{1}{16}=\frac{15}{16} \\ \cos \theta =\pm \frac{\sqrt{15}}{4} \\ \cos \theta =-\frac{\sqrt{15}}{4} \end{gathered}$$

Part (b) Step 1. Finding tanθ

$$\begin{gathered} \tan \theta =\frac{1/4}{-\sqrt{15}/4} \\ \tan \theta =\frac{1}{4}\times \frac{-4}{\sqrt{15}} \\ \tan \theta =-\frac{1}{\sqrt{15}} \end{gathered}$$

Part (c) Step 1. Finding sin2θ

$$\begin{gathered} \sin \left(2\theta \right)=2\times \frac{1}{4}\times -\frac{\sqrt{15}}{4} \\ \sin \left(2\theta \right)=-\frac{\sqrt{15}}{8} \end{gathered}$$

Part (d) Step 1. Finding cos2θ

$$\begin{gathered} \cos 2\theta ={\left(-\frac{\sqrt{15}}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2 \\ \cos 2\theta =\frac{15}{16}-\frac{1}{16}=\frac{14}{16}=\frac{7}{8} \end{gathered}$$

Part (e) Step 1. Finding sin12θ

$$\begin{gathered} \sin \left(\frac{1}{2}\theta \right)=\sqrt{\frac{1-\left(-{\displaystyle \raisebox{1ex}{$\sqrt{15}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}{2}} \\ \sin \left(\frac{1}{2}\theta \right)=\sqrt{\frac{1+\raisebox{1ex}{$\sqrt{15}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}} \\ \sin \left(\frac{1}{2}\theta \right)=\sqrt{\frac{4+\sqrt{15}}{8}} \\ \sin \left(\frac{1}{2}\theta \right)=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}} \end{gathered}$$