Question 4, Exercise 8.1
Solutions of Question 4 of Exercise 8.1 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.
Question 4(i)
Rewrite as a single expression. $\cos 6 \theta \cos 3 \theta-\sin 6 \theta \sin 3 \theta$
Solution.
\begin{align*} & \cos 6 \theta \cos 3 \theta-\sin 6 \theta \sin 3 \theta \\ & = \cos (6\theta +3\theta) \\ & = \cos 9\theta . \end{align*}
Question 4(ii)
Rewrite as a single expression. $\cos 7 \theta \cos 2 \theta+\sin 7 \theta \sin 2 \theta$.
Solution.
\begin{align*}
& \cos 7 \theta \cos 2 \theta + \sin 7 \theta \sin 2 \theta \\
& = \cos (7\theta -2\theta) \\
& = \cos 5\theta .
\end{align*}
Question 4(iii)
Rewrite as a single expression. $\sin \left(\frac{\theta}{3}\right) \cos \left(\frac{\theta}{6}\right)+\cos \left(\frac{\theta}{3}\right) \sin \left(\frac{\theta}{6}\right)$
Solution.
\begin{align*}
& \sin \left(\frac{\theta}{3}\right) \cos \left(\frac{\theta}{6}\right) + \cos \left(\frac{\theta}{3}\right) \sin \left(\frac{\theta}{6}\right) \\
& = \sin \left(\frac{\theta}{3} + \frac{\theta}{6}\right) \\
& = \sin \left(\frac{2\theta}{6} + \frac{\theta}{6}\right) \\
& = \sin \left(\frac{3\theta}{6}\right) \\
& = \sin \left(\frac{\theta}{2}\right)
\end{align*}
Question 4(iv)
Rewrite as a single expression. $\sin 138^{\circ} \cos 46^{\circ}-\cos 138^{\circ} \sin 46^{\circ}$.
Solution.
\begin{align*} & \sin 138^{\circ} \cos 46^{\circ} - \cos 138^{\circ} \sin 46^{\circ} \\ & = \sin(138^{\circ} - 46^{\circ}) \\ & = \sin(92^{\circ}). \end{align*}
Question 4(v)
Rewrite as a single expression. $\dfrac{\tan 75^{\circ}-\tan 45^{\circ}}{1+\tan 75^{\circ} \tan 45^{\circ}}$
Solution.
\begin{align*} & \dfrac{\tan 75^{\circ} - \tan 45^{\circ}}{1 + \tan 75^{\circ} \tan 45^{\circ}} \\ & = \tan(75^{\circ} - 45^{\circ}) \\ & = \tan(30^{\circ}). \end{align*}
Question 4(vi)
Rewrite as a single expression. $\dfrac{\tan \frac{4 \pi}{3}+\tan \frac{2 \pi}{3}}{1-\tan \frac{4 \pi}{3} \tan \frac{2 \pi}{3}}$
Solution.
\begin{align*} & \dfrac{\tan \frac{4 \pi}{3} + \tan \frac{2 \pi}{3}}{1 - \tan \frac{4 \pi}{3} \tan \frac{2 \pi}{3}} & = \tan\left(\frac{4\pi}{3} + \frac{2\pi}{3}\right) \\ & = \tan\left(2\pi\right). \end{align*}
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