Question 1, 2 and 3 Exercise 8.2

Solutions of Question 1, 2 and 3 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Suppose $P(-3,4)$ lies on the terminal side of $\theta$ when $\theta$ is plotted in standard position. Find $\cos 2 \theta$ and $\sin 2 \theta$ and determine the quadrant in which the terminal side of the angle $2 \theta$ lies when it is plotted in standard position.

Solution.

Given: $x=-3$ and $y=4$.

\begin{align*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5. \end{align*} Thus $$\sin\theta = \frac{4}{5} \text{ and } \cos\theta = -\frac{3}{5}.$$ Now \begin{align*} \sin2\theta&= 2\sin\theta \cos\theta\\ &= 2\left(\frac{4}{5} \right) \left(-\frac{3}{5} \right) \\ & = -\frac{24}{25}. \end{align*} and \begin{align*} \cos2\theta&= 1-2\sin^2\theta\\ &= 1-2\left(\frac{4}{5} \right)^2 \\ & = 1-\frac{32}{25} \\ & = -\frac{7}{25} \end{align*} Since $\sin2\theta$ and $\cos 2\theta$ both are negative. This gives $2\theta$ lies in III quadrant.

If $\sin \alpha=y$ and $\alpha$ lies in QII. Find expressions for $\sin 2 \alpha$, $\cos 2 \alpha$ and $\tan 2 \alpha$ in terms of $y$.

Solution.

Given: $\sin \alpha=y$ and $\alpha$ lies in QII.

We have an identity: $$\cos \alpha = \pm \sqrt{1-\sin^2 \alpha}$$ Since $\alpha$ lies in QII and $\cos$ is negative in QII, therefore \begin{align*} \cos \alpha & = - \sqrt{1-\sin^2 \alpha} \\ & =- \sqrt{1-y^2} \end{align*}. Now \begin{align*} & \sin 2\alpha = 2 \sin \alpha \cos \alpha \\ \implies & \boxed{\sin 2\alpha=- y\sqrt{1-y^2}} \end{align*} \begin{align*} & \cos 2\alpha = 1-2\sin^2 \alpha \\ \implies & \boxed{\cos 2\alpha=1-2y^2} \end{align*} \begin{align*} & \tan 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} \\ \implies & \boxed{\tan 2\alpha=-\frac{y\sqrt{1-y^2}}{1-2y^2}} \end{align*}

Use a half angle formula to find the exact value of $\cos 15^{\circ}$.

Solution.

\begin{align*} \sin 15^{\circ}&=\sin \left(\frac{30^{\circ}}{2}\right)\\ &=\sqrt{1-\cos30^{\circ}}{2}\\ &=\sqrt{\frac{1-0.866}{2}}\\ &=\sqrt{\frac{0.134}{2}}=\sqrt{0.067}=0.259 \end{align*}

Question 4 >