Trigonometric Equations Question 243
Question: If in $ \Delta ,ABC $ , $ 2b^{2}=a^{2}+c^{2}, $ then $ \frac{\sin 3B}{\sin B}= $
[UPSEAT 1999]
Options:
A) $ \frac{c^{2}-a^{2}}{2ca} $
B) $ \frac{c^{2}-a^{2}}{ca} $
C) $ {{( \frac{c^{2}-a^{2}}{ca} )}^{2}} $
D) $ {{( \frac{c^{2}-a^{2}}{2ca} )}^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \frac{\sin 3B}{\sin B}=\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=3-4{{\sin }^{2}}B $ $ =3-4+4{{\cos }^{2}}B=-1+\frac{4{{(a^{2}+c^{2}-b^{2})}^{2}}}{4{{(ac)}^{2}}} $ $ =-1+\frac{{{( \frac{a^{2}+c^{2}}{2} )}^{2}}}{{{(ac)}^{2}}}=-1+\frac{{{(a^{2}+c^{2})}^{2}}}{4{{(ac)}^{2}}} $ $ =\frac{{{(a^{2}+c^{2})}^{2}}-4a^{2}c^{2}}{4{{(ac)}^{2}}}={{( \frac{c^{2}-a^{2}}{2ac} )}^{2}} $ .
BETA
