====== 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.

=====Question 1(i)=====
Express the product as sum or difference $2\sin 6x\sin x$.
====Solution==== 
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}


=====Question 1(ii)=====
Express the product as sum or difference $\sin {{55}^{\circ }}\cos {{123}^{\circ }}$.
====Solution==== 
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}

=====Question 1(iii)=====
Express the product as sum or difference: $$\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}.$$
====Solution==== 
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}
$$\implies \sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}=\dfrac{1}{2}\left( \sin A+\sin B \right).$$

=====Question 1(iv)=====
Express the product as sum or difference: $$\cos \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}.$$
====Solution==== 
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}
$$\implies\sin \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}=\dfrac{1}{2}\left( \cos P-\cos Q \right)$$
====Go to====
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