# Question 2, Exercise 10.3

Solutions of Question 2 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.

Convert the sum or difference as product: $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}.$$

We have an identity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{\circ }}$, $\beta ={{43}^{\circ }}$ \begin{align}\sin {{37}^{\circ }}+\sin {{43}^{\circ }}&=2\sin \left( \dfrac{{{37}^{\circ }}+{{43}^{\circ }}}{2} \right)\cos \left( \dfrac{{{37}^{\circ }}-{{43}^{\circ }}}{2} \right)\\ &=2\sin \left( \dfrac{{{80}^{\circ }}}{2} \right)\cos \left( \dfrac{-{{6}^{\circ }}}{2} \right)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=2\sin {{40}^{\circ }}\cos {{3}^{\circ }}.$$

Convert the sum or difference as product $\cos {{36}^{\circ }}-\cos {{82}^{\circ }}$.

We have an identity: $$\cos \alpha -\cos \beta =-2\sin \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\alpha -\beta }{2} \right)$$ Put $\alpha =36^\circ$, adn $\beta =82^\circ$ \begin{align}\cos {{36}^{\circ }}-\cos {{82}^{\circ }}&=-2\sin \left( \dfrac{{{36}^{\circ }}+{{82}^{\circ }}}{2} \right)\sin \left( \dfrac{{{36}^{\circ }}-{{82}^{\circ }}}{2} \right)\\ &=-2\sin \left( \dfrac{{{118}^{\circ }}}{2} \right)\sin \left( \dfrac{-{{46}^{\circ }}}{2} \right).\\ &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{36}^{\circ }}-\cos {{82}^{\circ }}=2\sin 59^\circ \sin23^\circ.$$

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

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

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

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