====== 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====
\begin{align}L.H.S&=\cos \alpha \\
\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)\\
\cos \alpha &=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\\
\cos \alpha & =1-2{{\sin }^{2}}\dfrac{\alpha }{2}\\
\cos \alpha &=2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}\end{align}
=====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}
[[fsc-part1-kpk:sol:unit10:ex10-1-p5|< Question 5]]
[[fsc-part1-kpk:sol:unit10:ex10-1-p7|Question 7 >]]