Quantitative Aptitude Ques 2343

Question: If $ \sin \theta +\cos \theta =x, $ then the value of $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta $ is equal to

Options:

A) $ \frac{1}{4} $

B) $ \frac{1}{4}(1+6x^{2}) $

C) $ \frac{1}{4}(1+6x^{2}-3x^{4}) $

D) $ \frac{1}{2}(5-3x^{2}) $

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Answer:

Correct Answer: C

Solution:

  • $ \sin \theta +\cos \theta =x $ On squaring both sides, we get $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =x^{2} $ $ 1+2\sin \theta \cos \theta =x^{2} $

$ \therefore $ $ \sin \theta \cos \theta =\frac{x^{2}-1}{2} $ $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\cos }^{2}}\theta )}^{3}}+{{({{\sin }^{2}}\theta )}^{3}} $ $ =({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )(cos^{4}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{4}}\theta ) $ $ =[{{(cos^{2}\theta )}^{2}}+{{({{\sin }^{2}}\theta )}^{2}}-{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =[{{(cos^{2}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta =1-3{{( \frac{x^{2}-1}{2} )}^{2}} $ $ =1-\frac{3(x^{4}-2x^{2}+1)}{4}=\frac{4-3x^{4}+6x^{2}-3}{4} $ $ =\frac{1-3x^{4}+6x^{2}}{4}=\frac{1}{4}(1+6x^{2}-3x^{4}) $