Trigonometric Identities Question 250
Question: If $ \sin A,(60{}^\circ -A),\sin (60{}^\circ +A)=k\sin 3A, $ then what is k equal to?
Options:
A) $ 1/4 $
B) $ 1/2 $
C) $ 1 $
D) $ 4 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sin A.\sin (60{}^\circ -A)sin(60{}^\circ +A)=k,sin3A $
$ \Rightarrow ,\sin A.\frac{\sin 3A}{4\sin A}=k.\sin 3A $ $ [ \because ,\sin (60{}^\circ +A).sin(60{}^\circ -A)=\frac{\sin 3A}{4\sin A} ] $
$ \Rightarrow ,\frac{\sin 3A}{4}=k.\sin 3A $
$ \therefore ,k=\frac{1}{4} $
BETA
