A) \[\frac{(1-cosA)}{2}\]
B) \[\frac{(1+cosA)}{2}\]
C) \[\frac{(1+\operatorname{sinA})}{2}\]
D) \[\frac{(1-\operatorname{sinA})}{2}\]
Correct Answer: B
Solution :
\[\cos A={{\cos }^{2}}\frac{A}{2}-{{\sin }^{2}}\frac{A}{2}\] \[\Rightarrow \]\[\cos A={{\cos }^{2}}\frac{A}{2}-\left( 1-{{\cos }^{2}}\frac{A}{2} \right)\] \[\Rightarrow \]\[\cos A=2{{\cos }^{2}}\frac{A}{2}-1\] \[\Rightarrow \]\[2{{\cos }^{2}}\frac{A}{2}=1+\cos A\] \[\Rightarrow \]\[{{\cos }^{2}}\frac{A}{2}=\frac{1+\cos A}{2}\] Or Put \[A={{60}^{o}}\] L.H.S \[={{\cos }^{2}}\frac{60}{2}={{\cos }^{2}}{{30}^{o}}\] \[={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{3}{4}\] RHS\[=\frac{1+\cos 60}{2}=\frac{1+\frac{1}{2}}{2}=\frac{3}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
