| If \[\sin \theta +\cos \theta =x,\] then the value of \[{{\cos }^{6}}\theta +{{\sin }^{6}}\theta \]is equal to |
A) \[\frac{1}{4}\]
B) \[\frac{1}{4}(1+6{{x}^{2}})\]
C) \[\frac{1}{4}(1+6{{x}^{2}}-3{{x}^{4}})\]
D) \[\frac{1}{2}(5-3{{x}^{2}})\]
Correct Answer: C
Solution :
| \[\sin \theta +\cos \theta =x\] |
| On squaring both sides, we get |
| \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{x}^{2}}\] |
| \[1+2\sin \theta \cos \theta ={{x}^{2}}\] |
| \[\therefore \] \[\sin \theta \cos \theta =\frac{{{x}^{2}}-1}{2}\] |
| \[{{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\cos }^{2}}\theta )}^{3}}+{{({{\sin }^{2}}\theta )}^{3}}\] |
| \[=({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )(co{{s}^{4}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{4}}\theta )\] |
| \[=[{{(co{{s}^{2}}\theta )}^{2}}+{{({{\sin }^{2}}\theta )}^{2}}-{{\cos }^{2}}\theta {{\sin }^{2}}\theta ]\] |
| \[=[{{(co{{s}^{2}}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta ]\]\[=1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta =1-3{{\left( \frac{{{x}^{2}}-1}{2} \right)}^{2}}\] |
| \[=1-\frac{3({{x}^{4}}-2{{x}^{2}}+1)}{4}\,\,=\,\,\frac{4-3{{x}^{4}}+6{{x}^{2}}-3}{4}\] |
| \[=\,\,\frac{1-3{{x}^{4}}+6{{x}^{2}}}{4}=\frac{1}{4}(1+6{{x}^{2}}-3{{x}^{4}})\] |
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