Question 1, Exercise 10.1

Solutions of Question 1 of Exercise 10.1 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. There are four parts in Question 1.

Write as a trigonometric function of a single angle. $\sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$

As $$\begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align}$$ Therefore $$\begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }} & =\sin \left( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align}$$

Write as a trigonometric function of a single angle. $\cos {{83}^{\circ }}\cos {{53}^{\circ }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}$

As $$\begin{align}\cos (\alpha -\beta )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta,\end{align}$$ Therefore $$\begin{align}\cos {{83}^{\circ }}\cos {{53}^{\circ }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}&=\cos \left( 83-53 \right)\\ &=\cos {{30}^{\circ }}\end{align}$$ .

Write as a trigonometric function of a single angle. $\cos {{19}^{\circ }}\cos {{5}^{\circ }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}$

As $$\begin{align}\cos (\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta,\end{align}$$ Therefore $$\begin{align}\cos {{19}^{\circ }}\cos {{5}^{\circ }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}&=\cos \left( 19+5 \right)\\ &=\cos {{24}^{\circ }}\end{align}$$ .

Write as a trigonometric function of a single angle. $\sin {{40}^{\circ }}\cos {{15}^{\circ }}-\cos {{40}^{\circ }}\sin {{15}^{\circ }}$

As $$\begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta,\end{align}$$ Therefore $$\begin{align}\sin {{40}^{\circ }}\cos {{15}^{\circ }}-\cos {{40}^{\circ }}\sin {{15}^{\circ }}&=\sin \left( 40-15 \right)\\ &=\sin {{25}^{\circ }}\end{align}$$ .

Write as a trigonometric function of a single angle. $\dfrac{\tan {{20}^{\circ }}+\tan {{32}^{\circ }}}{1-\tan {{20}^{\circ }}\tan {{32}^{\circ }}}$

As $$\begin{align}\tan (\alpha +\beta )=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta },\end{align}$$ Therefore $$\begin{align}\frac{\tan {{20}^{\circ }}+\tan {{32}^{\circ }}}{1-\tan {{20}^{\circ }}\tan {{32}^{\circ }}}&=\tan \left( 20+32 \right)\\ &=\tan {{52}^{\circ }}.\end{align}$$

Write as a trigonometric function of a single angle. $\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}$

As $$\begin{align}\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta },\end{align}$$ Therefore $$\begin{align}\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}&=\tan \left( 35-12 \right)\\ &=\tan {{23}^{\circ }}\end{align}$$ .

Question 2 >