| If \[\sin \theta \cos \theta =1/2,\]then what is \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta \]equal to? |
A) 1
B) 2
C) 3
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
| Given, \[\sin \theta .\cos \theta =\frac{1}{2}\] |
| \[{{\sin }^{6}}\theta +{{\cos }^{2}}\theta ={{({{\sin }^{2}}\theta )}^{3}}+{{({{\cos }^{2}}\theta )}^{3}}\] |
| \[=\,\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )({{\sin }^{4}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta )\] |
| \[[\because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] |
| \[={{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] |
| \[=(1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta )\] \[\left[ \because \sin \theta \cdot \cos \theta =\frac{1}{2} \right]\] |
| \[=1-3\times \frac{1}{4}=1-\frac{3}{4}=\frac{1}{4}\] |
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