Chapter 5, Problem 106RE

To determine

To calculate: The exact values of the sine, cosine and tangent of the given angle.

Answer to Problem 106RE

  $\begin{array}{l}\sin {157}^{\text{o}}30\text{'}=\frac{1}{2}\sqrt{2-\sqrt{2}}\\ \cos {157}^{\text{o}}30\text{'}=\frac{-1}{2}\sqrt{2+\sqrt{2}}\\ \tan {157}^{\text{o}}30\text{'}=-(\sqrt{2}-1)\end{array}$

Explanation of Solution

Given information:

The given angle:

  ${157}^{\text{o}}30\text{'}$

Formula used:

Half-Angle Formulas:

  $\begin{array}{l}\sin \frac{u}{2}=\pm \sqrt{\frac{1-\cos u}{2}}\\ \cos \frac{u}{2}=\pm \sqrt{\frac{1+\cos u}{2}}\\ \tan \frac{u}{2}=\frac{1-\cos u}{\sin u}=\frac{\sin u}{1+\cos u}\end{array}$

The signs of $\sin \frac{u}{2}$ and $\cos \frac{u}{2}$ depends on the quadrant in which $\frac{u}{2}$ lies.

Calculation:

As per problem,

The given angle:

  ${157}^{\text{o}}30\text{'}$

Note the ${157}^{\text{o}}30\text{'}$ is half of ${315}^{\text{o}}$

1. Use half-angle formula for $\sin \frac{u}{2}$ and the fact that ${157}^{\text{o}}30\text{'}$ lies in Quadrant II.

  $\begin{array}{l}\sin {157}^{\text{o}}30\text{'}=\sqrt{\frac{1-\cos {315}^{\text{o}}}{2}}\\ =\sqrt{\frac{1-(\cos {45}^{\text{o}})}{2}}\text{ [cos(}{360}^{\text{o}}-\theta )=\cos \theta ;({315}^{\text{o}}={360}^{\text{o}}-{45}^{\text{o}})]\\ =\sqrt{\frac{1-(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.)}{2}}\\ =\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\ =\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}\cdot \frac{2\sqrt{2}}{2\sqrt{2}}}\text{ [Multiply both numerator and denominator by }2\sqrt{2}\text{ to rationalise]}\\ =\sqrt{\frac{(\sqrt{2}-1)(2\sqrt{2})}{8}}\\ =\sqrt{\frac{(2-\sqrt{2})}{4}}\text{ [Divide out common factors and multiply]}\\ =\frac{1}{2}\sqrt{2-\sqrt{2}}\end{array}$

The positive square root is chosen because sine of an angle is positive in Quadrant II.

2. Use half-angle formula for $\cos \frac{u}{2}$ and the fact that ${157}^{\text{o}}30\text{'}$ lies in Quadrant II.

  $\begin{array}{l}\cos {157}^{\text{o}}30\text{'}=-\sqrt{\frac{1+\cos {315}^{\text{o}}}{2}}\\ =-\sqrt{\frac{1+(\cos {45}^{\text{o}})}{2}}\text{ [cos(}{360}^{\text{o}}-\theta )=\cos \theta ;({315}^{\text{o}}={360}^{\text{o}}-{45}^{\text{o}})]\\ =-\sqrt{\frac{1+(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.)}{2}}\\ =-\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}\\ =-\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}\cdot \frac{2\sqrt{2}}{2\sqrt{2}}}\text{ [Multiply both numerator and denominator by }2\sqrt{2}\text{ to rationalise]}\\ =-\sqrt{\frac{(\sqrt{2}+1)(2\sqrt{2})}{8}}\\ =-\sqrt{\frac{(2+\sqrt{2})}{4}}\text{ [Divide out common factors and multiply]}\\ =\frac{-1}{2}\sqrt{2+\sqrt{2}}\end{array}$

The negative root is chosen because cosine of an angle is negative in Quadrant II.

3. Use half-angle formula for $\tan \frac{u}{2}$ and the fact that ${157}^{\text{o}}30\text{'}$ lies in Quadrant II.

  $\begin{array}{l}\tan {157}^{\text{o}}30\text{'}=\frac{1-\cos {315}^{\text{o}}}{\sin {315}^{\text{o}}}\\ =\frac{1-\cos {45}^{\text{o}}}{-\sin {45}^{\text{o}}}\text{ [cos(}{360}^{\text{o}}-\theta )=\cos \theta \text{ and }\sin ({360}^{\text{o}}-\theta )=-\sin \theta ;({315}^{\text{o}}={360}^{\text{o}}-{45}^{\text{o}})]\\ =\frac{1-(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.)}{-(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.)}\\ =-(\sqrt{2}-1)\text{ [Simplify]}\end{array}$

Hence,

  $\begin{array}{l}\sin {157}^{\text{o}}30\text{'}=\frac{1}{2}\sqrt{2-\sqrt{2}}\\ \cos {157}^{\text{o}}30\text{'}=\frac{-1}{2}\sqrt{2+\sqrt{2}}\\ \tan {157}^{\text{o}}30\text{'}=-(\sqrt{2}-1)\end{array}$