A) 1
B) \[co{{s}^{3}}x\text{ }si{{n}^{3}}x\]
C) 0
D) \[\infty \]
Correct Answer: A
Solution :
Given \[sin\text{ }x+sin\text{ }x=1\] \[\Rightarrow \] \[\sin x=1-{{\sin }^{2}}x={{\cos }^{2}}x\] \[\Rightarrow \] \[\sin x={{\cos }^{2}}x\] \[\therefore \] \[{{\cos }^{6}}x+{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x\] \[=si{{n}^{3}}x+si{{n}^{6}}x+3\text{ }si{{n}^{5}}x+3\text{ }si{{n}^{4}}x\] \[={{(\sin x+{{\sin }^{2}}x)}^{3}}={{1}^{3}}=1\]
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