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- Question 1, Exercise 10.3
- 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... ut $\alpha =6x$ and $\beta =x$ \begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\\ &=\cos 7x
- Question 1, Exercise 10.2
- drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q1.png?nolink |reference triangle}} we find $\sin \theta =\dfrac{1}{\sqrt{26}}$ and $\cos \theta =\dfrac{-5}{\sqrt{26}}$ Thus, we have the following by using double angle identities. \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\left( \d
- Question 1, Exercise 10.1
- KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. ===== Question 1(i) ===== Write ... on of a single angle. $\sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\al
- Question 3, Exercise 10.2
- drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\c... drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\c... gle identities: \begin{align}\cos \dfrac{\theta }{2}&=\sqrt{\dfrac{1+\cos \theta }{2}}\\ &=\sqrt{\dfrac{1-\dfrac{3}{5}}{2}}=\sqrt{\dfrac{2}{10}}\end{ali