Trigonometric Functions And Graphs

The key is to simplify the quartic combination into a single low-order cosine term, because the period of a sum of trigonometric powers is dictated by the highest multiple-angle that survives reduction, not by the original functions. Begin with the algebraic identity ${\sin }^{4}x+{\cos }^{4}x=({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x$, where the first bracket is one, leaving $1-2{\sin }^{2}x{\cos }^{2}x$. Recognizing $2\sin x\cos x=\sin 2x$, the term $2{\sin }^{2}x{\cos }^{2}x=\frac{1}{2}{\sin }^{2}2x$, so the function is $1-\frac{1}{2}{\sin }^{2}2x$. Applying power reduction once more with ${\sin }^{2}2x=\frac{1-\cos 4x}{2}$ gives $1-\frac{1}{4}(1-\cos 4x)=\frac{3}{4}+\frac{1}{4}\cos 4x$. The periodicity is now governed entirely by the $\cos 4x$ term, whose fundamental period is $\frac{2\pi }{4}=\frac{\pi }{2}$. Option $\pi$ would correspond to a surviving $\cos 2x$ term, which does not appear after full reduction. Option $2\pi$ is the period of the unreduced sine and cosine. Option $\frac{\pi }{4}$ erroneously doubles the angular frequency. This follows the standard power-reduction periodicity pattern of JEE Advanced. As a final check, $g(0)=1$ and $g\left(\frac{\pi }{2}\right)=0+1=1$, and the function values repeat at spacing $\frac{\pi }{2}$, confirming the fundamental period.