Question

How to find the range of \[{\sin ^2}x + {\cos ^4}x\]

Answer 1

Hint: We have to derivation of the finding the range of the required trigonometric relation. At first, we will try to convert the given trigonometric function into a simplest for by using the trigonometric relations. Using the range of sine and cosine terms, we will get the final answer. From there we can easily find the range of the given function.

Formula used: As we know that, the range of \[\sin x\] and \[\cos x\] is \[[ - 1,1]\] and also we know that, \[\sin 2x = 2\sin x\cos x\]
Using this formula, we will convert \[{\sin ^2}x + {\cos ^4}x\] in terms of \[\sin x\] or \[\cos x\] function.

Complete step-by-step answer:
We have to find the range of \[{\sin ^2}x + {\cos ^4}x\].
We already know that the range of \[\sin x\] is \[[ - 1,1]\].
The range of \[\cos x\] is \[[ - 1,1]\].
Let us take, \[f(x) = {\sin ^2}x + {\cos ^4}x\]
Simplifying we get,
$\Rightarrow$\[f(x) = 1 - {\cos ^2}x + {\cos ^4}x\]
Simplifying again we get,
$\Rightarrow$\[f(x) = 1 - {\cos ^2}x(1 - {\cos ^2}x)\]
Since, we know that, \[{\sin ^2}x + {\cos ^2}x = 1\]
Simplifying again we get,
$\Rightarrow$\[f(x) = 1 - {\sin ^2}x{\cos ^2}x\]
Simplifying again we get,
$\Rightarrow$\[f(x) = 1 - \dfrac{{4{{\sin }^2}x{{\cos }^2}x}}{4}\]
Simplifying again we get,
$\Rightarrow$\[f(x) = 1 - \dfrac{{{{\sin }^2}2x}}{4}\]..................… (1)
Since, we know that, \[\sin 2x = 2\sin x\cos x\]
We have,
\[ - 1 \leqslant \sin 2x \leqslant 1\]
Squaring the terms, we get,
$\Rightarrow$\[0 \leqslant {\sin ^2}2x \leqslant 1\]
Dividing all the terms by 4 we get,
$\Rightarrow$\[0 \leqslant \dfrac{{{{\sin }^2}2x}}{4} \leqslant \dfrac{1}{4}\]
Multiplying all the terms by -1 we get,
$\Rightarrow$\[0 \geqslant - \dfrac{{{{\sin }^2}2x}}{4} \geqslant - \dfrac{1}{4}\]
Adding all the terms by 1 we get,
$\Rightarrow$\[1 - \dfrac{1}{4} \leqslant 1 - \dfrac{{{{\sin }^2}2x}}{4} \leqslant 1 - 0\]
Simplifying we get,
$\Rightarrow$\[\dfrac{3}{4} \leqslant 1 - \dfrac{{{{\sin }^2}2x}}{4} \leqslant 1\]
From 1 we know that $1 - \dfrac{{{{\sin }^2}2x}}{4}$ is equal to ${\sin ^2}x + {\cos ^4}x$.

$\therefore $ The range of \[{\sin ^2}x + {\cos ^4}x\] is \[\left[ {\dfrac{3}{4},1} \right]\].

Note: We have to remember that, the domain of any function is the set of all possible inputs (x – values) of a function whereas the range of any function is the set of all possible respective outcomes (y – values) of the function.
Domain lies on the X – axis and range lies on Y – axis.
For example, the function \[f(x) = \sin x\] has all real numbers in its domain, but its range is \[ - 1 \leqslant \sin x \leqslant 1\] .
The values of the sine function are different, depending on whether the angle is in degrees or radians. The function is periodic with periodicity ${360^ \circ }$ or \[2\pi \] radians.