A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[2\pi \]
D) None of these
Correct Answer: B
Solution :
Since, \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{1}{2}{{(2\sin x\cos x)}^{2}}\] \[=1-\frac{1}{2}{{(\sin 2x)}^{2}}=1-\frac{1}{2}\left( \frac{1-\cos 4x}{2} \right)\] \[=\frac{3}{4}+\frac{1}{4}\cos 4x\] Since, \[\cos x\] is periodic with period \[2\pi \]. \[\therefore \] The period of \[f(x)=\frac{2\pi }{4}=\frac{\pi }{2}\].
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