The period of sinπx2+2cosπx3−tanπx4

f(x)=sinπx2+2cosπx3−tanπx4 period of sinπx2=2ππ2=4, period of cosπx3=2ππ3=6, period of tanπx4=ππ4=4 ⇒Period = LCM of 4,6,4=12

The period of sinπx2+2cosπx3−tanπx4

f(x)=sinπx2+2cosπx3−tanπx4 period of sinπx2=2ππ2=4, period of cosπx3=2ππ3=6, period of tanπx4=ππ4=4 ⇒Period = LCM of 4,6,4=12

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The period of $\sin (\frac{π x}{2}) + 2 \cos (\frac{π x}{3}) - \tan (\frac{π x}{4})$

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Detailed Solution

$f (x) = \sin (\frac{πx}{2}) + 2 \cos (\frac{πx}{3}) - \tan (\frac{πx}{4}) period of \sin (\frac{πx}{2}) = \frac{2 π}{\frac{π}{2}} = 4, period of \cos (\frac{πx}{3}) = \frac{2 π}{\frac{π}{3}} = 6, period of \tan \frac{πx}{4} = \frac{π}{\frac{π}{4}} = 4 \Rightarrow Period = LCM of \{4, 6, 4\} = 12$