====== 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} 

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