Trigonometric Identities Question 301
Question: If $ \sin \theta +\cos \theta =x, $ then $ {{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\frac{1}{4}[4-3{{(x^{2}-1)}^{2}}] $ for
Options:
A) All real x
B) $ x^{2}\le 2 $
C) $ x^{2}\ge 2 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
On squaring the given relation $ \sin 2\theta =x^{2}-1\le 1\Rightarrow x^{2}\le 2 $ or $ -\sqrt{2}\le x\le \sqrt{2} $ $ [\because \sin 2\theta \le 1] $ Now $ {{\sin }^{6}}\theta +{{\cos }^{6}}\theta $ $ ={{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta ) $ $ =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1-\frac{3}{4}{{\sin }^{2}}2\theta $ $ =1-\frac{3}{4}{{(x^{2}-1)}^{2}}=\frac{1}{4}{4-3{{(x^{2}-1)}^{2}}} $ Thus the given result will hold true only when $ x^{2}\le 2 $ and not for all real values of x.
BETA
