Question

Find the range of f(x) = cos(sinx).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = cos(sinx)$

We know,

$\rm \: - 1 \: \leqslant \: sinx \: \leqslant \: 1$

As we know,

$\rm \: cos( - x) = cosx$

$\rm\implies \:cos(sinx) \: \sub \: [0, \: 1]$

Now,

$\rm \: f(x) = cos(sinx)$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx} f(x) =\dfrac{d}{dx} cos(sinx)$

$\rm \: f'(x) = - sin(sinx)\dfrac{d}{dx}sinx$

$\rm \: f'(x) = - sin(sinx) \: cosx$

$\rm\implies \:f'(x) \geqslant 0 \: \: if \: \: sinx \: \in \: [ - 1,0]$

$\rm\implies \:f(x) \: is \: increasing \: function \:$

So, it implies f(x) attains its minimum value at sinx = - 1

So, minimum value of f(x) = cos(- 1) = cos1

$\rm\implies \:f'(x) \leqslant 0 \: \: if \: \: sinx \: \in \: [0,1]$

$\rm\implies \:f(x) \: is \: decreasing \: function \:$

So, it implies, f(x) attains its maximum value at sinx = 0

So, maximum value of f(x) = cos0 = 1

Hence,

$\rm\implies \:Range \: of \: f(x) \: \in \: [cos1, \: 1]$

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BASIC CONCEPT USED

A function f(x) is said to be increasing if f(a) ≥ f(b) for every a > b

and

A function f(x) is said to be decreasing if f(a) ≤ f(b) for every a > b

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$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$

Answer 2

Question:-

Find the range of f(x) = cos(sin x).

Given:-

• The Function f(x) = cos(sin x).

To Find:-

• The Range of the Function.

Solution:-

$\sf \large Let, f:\mathbb{R}</p><p> \longmapsto A,f(x) = cos(sin \: x)$

There are multiple ways to go about this. We can use the fact that ʄ is a composed function or do it directly. We'll use the second approach.

We know that the range of sinx and cosx is

 [ − 1, 1 ]. Hence, The range of cos(sinx) will be located in the interval [ − 1, 1 ]. However, the exact range, denoted A in this answer, is going to be:

A = [ cos(x₁), cos(x₂) ]

Where cos(x₁) is the minimal cosine value in the interval [ − 1, 1 ] and cos(x₂) is the maximum value. As cos(x) is periodic, so will cos(sinx).

On the interval [ − 1, 0] the cosine function is monotonic increasing, hence the minimal value of cosx is going to be when x = − 1. Similarly, as the cosine is monotic decreasing on [ 0, 1 ], this means 

x = 0 is the max value and it also means that x = 1 is also the minimal value.

∴ A = [ cos( – 1), cos 0 ] = [ cos 1, cos 0 ] = [ cos 1, 1 ].

Answer:-

${ \boxed{ \sf \large \green{Thus, \: the \: range \: is \: A = [ cos 1, 1 ].}}}$

Hope you have satisfied. ⚘