Period of $\sin^2 x +\cos^4 x$?

Question

How do I approach this problem? I tried reducing it to $\sin ax + \cos bx$ form but can't. I converted it to 1+ $\sin^2 x \cos^2 x $, but I don't know if I can get the answer from it. Would someone please help me with this? The answer is $\pi$.

Answer 1

A bit of trickery:

$\sin^2 x -\sin^2 x -\cos^2 x +1 + \cos^4 x$

$=\cos^2 x(\cos^2 x-1) +1=$

$ \cos^2 x\sin^2 x+1=$

$(1/4)\sin^2 (2x) +1$.

The fundamental period is?

Appended:

1) The fundamental period of $\sin y$ is $2π$.

2) The fundamental period of $\sin^2 y$ is $π$. (Why?)

(Formally :$ \cos (2y)=$ $ \cos^2 y - \sin^2 y=1- 2\sin^2 y$, or $\sin^2 y =(1/2)(1-\cos (2y))$

3) The fundamental period of $\sin^2 (2y)$ is $π/2$.(Why?)

Answer 2

$ = 1-\cos^2(x) +\cos^4(x) = 1 - \cos^2(x)(1-\cos^2(x)) = 1-(\cos(x)\sin(x))^2 = 1-(1/4)\cdot \sin^2(2x) =(1/8)\cdot(7+\cos(4x)) $

Answer 3

Since$$\sin\left(x+\frac\pi2\right)=\cos(x)\text{ and }\cos\left(x+\frac\pi2\right)=-\sin(x),$$it is clear that $\sin^2+\cos^4$ is periodic with period $\frac\pi2$. But is $\frac\pi2$ the smallest period? Yes, because\begin{align}(\sin^2+\cos^4)'(x)&=2\sin(x)\cos(x)-4\cos^3(x)\sin(x)\\&=2\sin(x)\cos(x)\times\bigl(1-2\cos^2(x)\bigr)\\&=2\sin(x)\cos(x)\times\bigl(\sin^2(x)-\cos^2(x)\bigr)\\&=-\sin(2x)\cos(2x)\\&=-\frac{\sin(4x)}2\end{align}and so obviously $\frac\pi2$ is the smallest period of $(\sin^2+\cos^4)'$.

Answer 4

We have that

$$\sin^2 x +\cos^4 x=\frac12(1-\cos 2x)+\frac14(1+\cos 2x)^2=$$

$$=\frac12-\frac12 \cos 2x+\frac14+\frac12 \cos 2x+\frac14\cos^2 2x=$$

$$=\frac14 \cos^2 2x+\frac 34=\frac18 (1+\cos 4x)+\frac34=\frac18 \cos 4x+\frac78$$

therefore the period is equal to $\frac{\pi}2$.

Answer 5

Let $f(x)=\sin^2x+\cos^4x$. It's straightforward (if a bit tedious) to show that $f(x+\pi/2)=f(x)$, so the period of $f$ divides $\pi/2$. But for $0\lt x\lt\pi/2$ we have

$$f(x)=\sin^2x+\cos^4x\lt\sin^2x+\cos^2x=1=f(0)$$

so the period of $f$ cannot be less than $\pi/2$. Thus the period is $\pi/2$.