# Question 2, Exercise 10.2

Solutions of Question 2 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta$ is in the second quadrant, then find $\sin 2\theta$.

Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity $$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\ &=-\sqrt{1-\left(\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\dfrac{120}{169}.}$$

If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta$ is in the second quadrant, then find $\cos 2\theta$.

Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity $$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\ &=-\sqrt{1-\left(\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin }^{2}}\theta\\ &={{\left( \dfrac{12}{13} \right)}^{2}}-{{\left( \dfrac{-5}{13} \right)}^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos 2\theta=-\dfrac{119}{169}.}$$

If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta$ is in the second quadrant, then find $\tan 2\theta$.

Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity $$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\ &=-\sqrt{1-\left(\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: Thus, we have the following by using double angle identities. \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)=-\dfrac{120}{169}\end{align} and \begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\ &={{\left( \dfrac{12}{13} \right)}^{2}}-{{\left( \dfrac{-5}{13} \right)}^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}=\dfrac{119}{169}.\end{align} Now \begin{align}\tan 2\theta &=\dfrac{\sin 2\theta }{\cos 2\theta }\\ &=\dfrac{\frac{-120}{169}}{\frac{119}{169}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{120}{119}}$$