| If \[\sin \theta +\cos \theta =x,\] then the value of\[{{\cos }^{6}}\theta +{{\sin }^{6}}\theta \] is equal to [SSC (CGL) 2010] |
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 :
| \[\sin \theta +\cos \theta =x\] |
| \[{{\sin }^{2}}+{{\cos }^{2}}\theta +2\sin \theta \cos ={{x}^{2}}\] |
| \[\sin \theta \times \cos \theta =\frac{{{x}^{2}}-1}{2}\] |
| Now, \[{{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\sin }^{2}}\theta )}^{3}}+{{({{\cos }^{2}}\theta )}^{3}}\] |
| \[=({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )({{\cos }^{4}}\theta +{{\sin }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta )\] |
| \[={{({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]\[={{(co{{s}^{2}}\theta +{{\sin }^{2}}\theta )}^{2}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] |
| \[=1-3{{\left( \frac{{{x}^{2}}-1}{2} \right)}^{2}}=\frac{1}{4}(1+6{{x}^{2}}-3{{x}^{4}})\] |
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