Question 1, Exercise 10.3

Solutions of Question 1 of Exercise 10.3 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.

Express the product as sum or difference $2\sin 6x\sin x$.

We have an identity: $$-2\sin \alpha \sin \beta =\cos (\alpha +\beta )-\cos (\alpha -\beta ).$$ Put $\alpha =6x$ and $\beta =x$ $$\begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\\ &=\cos 7x-\cos x\\ 2\sin 6x\sin x&=\cos 5\theta -\cos 7\theta\end{align}$$

Express the product as sum or difference $\sin {{55}^{\circ }}\cos {{123}^{\circ }}$.

We have an identity: $$2\sin \alpha \cos \beta =\sin (\alpha +\beta )+\sin (\alpha -\beta )$$ Put $\alpha ={{55}^{\circ }}$ and $\beta ={{123}^{\circ }}$ \begin{align}2\sin {{55}^{\circ }}\cos {{123}^{\circ }}&=\sin ({{55}^{\circ }}+{{123}^{\circ }})+\sin ({{55}^{\circ }}-{{123}^{\circ }})\\ &=\sin {{178}^{\circ }}+\sin(-68^\circ)\\ &=\sin {{178}^{\circ }}-\sin {{68}^{\circ }}\\ \implies \sin {{55}^{\circ }}\cos {{123}^{\circ }}&=\dfrac{1}{2} \left[ \sin {{178}^{\circ }}-\sin {{68}^{\circ }} \right]. \end{align}

Express the product as sum or difference: $$\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}.$$

We have an identity: $$2\sin \alpha \cos \beta =\sin (\alpha +\beta )+\sin (\alpha -\beta )$$ Put $\alpha =\dfrac{A+B}{2}$ and $\beta =\dfrac{A-B}{2}$ $$\begin{align}& 2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}\\ &=\sin \left(\dfrac{A+B}{2}+\dfrac{A-B}{2}\right)+\sin \left(\dfrac{A+B}{2}-\dfrac{A-B}{2} \right)\\ &=\sin A+\sin B.\end{align}$$ $$\implies \sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}=\dfrac{1}{2}\left( \sin A+\sin B \right).$$

Express the product as sum or difference: $$\cos \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}.$$

We have an identity: $$2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$$ Put $\alpha =\dfrac{P+Q}{2}$ and $\beta =\dfrac{P-Q}{2}$ $$\begin{align}& 2\cos \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}\\ &=\cos \left(\dfrac{P+Q}{2}+\dfrac{P-Q}{2}\right)-\cos \left(\dfrac{P+Q}{2}-\dfrac{P-Q}{2}\right)\\ &=\cos P-\cos Q \end{align}$$ $$\implies\sin \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}=\dfrac{1}{2}\left( \cos P-\cos Q \right)$$

Question 2 >