Trigonometric Identities Question 365
Question: $ sinA+2sin,2A+sin3A $ is equal to which of the following?
- $ 4\sin 2A{{\cos }^{2}}( \frac{A}{2} ) $
- $ 2\sin 2A{{( \sin \frac{A}{2}+\cos \frac{A}{2} )}^{2}} $
- $ 8\sin A\cos A{{\cos }^{2}}( \frac{A}{2} ) $ Select the correct answer using the code given below:
Options:
A) 1 and 2 only
B) 2 and 3 only
C) 1 and 3 only
D) 1, 2 and 3
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ A=30{}^\circ $
$ \Rightarrow \sin A+2\sin 2A+\sin 3A $ $ =\sin 30{}^\circ +2\sin 60{}^\circ +\sin 90{}^\circ $ $ =\frac{1}{2}+\frac{2\sqrt{3}}{2}+1=\frac{2\sqrt{3}+3}{2} $ $ (\because 2{{\cos }^{2}}A=1+\cos 2A) $ Now, $ 4\sin 2A{{\cos }^{2}}( \frac{A}{2} )=2\sin 2A\ [1+\cos A] $ $ =2\sin 60{}^\circ [1+\cos 30{}^\circ ]=\frac{2\sqrt{3}+3}{2} $ Also, $ \sin 2A=2\sin ,A,\cos A $ & $ {{\sin }^{2}}A+{{\cos }^{2}}A=1 $ $ 2\sin 2A{{[ \sin \frac{A}{2}+\cos \frac{A}{2} ]}^{2}} $ $ =2\sin 2A[ {{\sin }^{2}}\frac{A}{2}+{{\cos }^{2}}\frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2} ] $ $ =2\sin 2A[ 1+\sin A ]=2\sin 60{}^\circ [ 1+\sin 30{}^\circ ]=\frac{3\sqrt{3}}{2} $ & $ 8\sin A\cos A,{{\cos }^{2}}( \frac{A}{2} ) $ $ =4\sin A\cos A[1+\cos A] $ $ =4\sin 30{}^\circ cos30{}^\circ [1+cos30{}^\circ ] $ $ =\frac{2\sqrt{3}+3}{2} $
BETA
