Question 6, 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.
Question 6(i)
Show that: $\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}$
Solution
We start from \begin{align}\cos \alpha &=\cos 2\dfrac{\alpha }{2}\\ &={{\cos }^{2}}\dfrac{\alpha }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}\\ &={{\cos }^{2}}\dfrac{\alpha }{2}-\left( 1-{{\cos }^{2}}\dfrac{\alpha }{2} \right)\\ &=2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \ldots (1)\end{align} Now \begin{align}2{{\cos }^{2}}\dfrac{\alpha }{2}-1&=2\left( 1-{{\sin }^{2}}\dfrac{\alpha }{2} \right)-1\\ &=2-2{{\sin }^{2}}\dfrac{\alpha }{2}-1\\ & =1-2{{\sin }^{2}}\dfrac{\alpha }{2} \ldots (2)\end{align} Now combining (1) and (2), we get $$\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}$$ as required.
Question 6(ii)
Show that: $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)={{\cos }^{2}}\beta -{{\cos }^{2}}\alpha$
Solution
\begin{align}L.H.S.&=\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)\\ &=\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right)\\ &={{\sin }^{2}}\alpha {{\cos }^{2}}\beta -{{\cos }^{2}}\alpha {{\sin }^{2}}\beta \\ &={{\sin }^{2}}\alpha \left( 1-si{{n}^{2}}\beta \right)-\left( 1-si{{n}^{2}}\alpha \right){{\sin }^{2}}\beta \\ &={{\sin }^{2}}\alpha -{{\sin }^{2}}\alpha si{{n}^{2}}\beta -{{\sin }^{2}}\beta +si{{n}^{2}}\alpha {{\sin }^{2}}\beta \\ &={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta \\ &=\left( 1-{{\cos }^{2}}\alpha \right)-\left( 1-{{\cos }^{2}}\beta \right)\\ &=1-{{\cos }^{2}}\alpha -1+{{\cos }^{2}}\beta \\ &={{\cos }^{2}}\beta -{{\cos }^{2}}\alpha \\ &=R.H.S.\end{align}