A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[2\pi \]
D) None of these
Correct Answer: B
Solution :
\[{{\sin }^{4}}x+{{\cos }^{4}}x={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}\] \[-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{1}{2}{{(\sin 2x)}^{2}}\] \[=1-\frac{1}{2}\left( \frac{1-\cos x4x}{2} \right)\] \[=1-\frac{1}{4}+\frac{1}{4}\cos 4x\] \[=\frac{3}{4}+\frac{1}{4}\cos 4x\] Since, period of\[cos\text{ }x\]is\[2\pi \]. \[\therefore \]Period of\[cos\text{ }4x\] is\[\frac{2\pi }{4}=\frac{\pi }{2}\] Hence, period of\[si{{n}^{4}}x+co{{s}^{4}}x\]is\[\frac{\pi }{2}\].
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