November 20, 2024

How to find the range of asinx+bcosx+c

by Bina singhMay 30, 2021October 25, 2023

How to find the range of asinx+bcosx+c ?

Since $sinx$ and $cosx$

is defined for all real values of $x$ , so the domain of the given function is the set of all real numbers. We have to find the range of asinx+bcosx+c . For the time being, assume that the quantity $\sqrt{a^{2}+b^{2}}$

is not zero ( if it was zero, it would then mean that both a and b are zero, resulting in f(x)=asinx+bcosx+c being a constant function of value c . In that case, the range would have been just c.

Range of asinx+bcosx

Maximum and minimum value of y=asinx+bcosx

To find the max. and min. value of asinx+bcosx , we will use the identity $sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta$ . We have $\large asinx+bcosx$

. So we would like to find an angle $\beta$ such that $cos\beta =a$

and

\sin β = b

, for then we could write $asinx+bcosx=cos\beta sinx+sin\beta cosx=sin(x+\beta )$

Since $sin\beta$ and $\cos β$

must be between

- 1

and

1

, and

a

and

b

may not be in that range. Moreover, we know that $cos^{2 }\beta +sin^{2}\beta$ must equal

1

, so we scale everything by $\sqrt{a^{2}+b^{2}}$

.

Let $A=\frac{a}{\sqrt{a^{2}+b^{2}}}$ and $B=\frac{b}{\sqrt{a^{2}+b^{2}}}$

. Clearly $A^{2}+B^{2}=1$ , so there is a unique angle $\beta$

such that $cos\beta =A$ and $sin\beta =B$

and $0\leq \beta < 2\pi$ . Then $asinx+bcosx =\sqrt{a^{2}+b^{2}}(Asinx+Bcosx)$

$=\sqrt{a^{2}+b^{2}}(cos\beta sinx+sin\beta cosx)$

=\sqrt{a^{2}+b^{2}}sin(x+\beta )

so $y=\sqrt{a^{2}+b^{2}}sin(x+\beta )$ .

Since $-1\leq sin(x+\beta )\leq 1$

, this implies

$=-\sqrt{a^{2}+b^{2}}\leq \sqrt{a^{2}+b^{2}}sin(x+\beta )\leq \sqrt{a^{2}+b^{2}}$

$\Rightarrow -\sqrt{a^{2}+b^{2}}\leq y\leq \sqrt{a^{2}+b^{2}}$

.

Range of asinx+bcosx+c

Let $y=f(x)=asinx+bcosx+c$ , then $y-c=asinx+bcosx$

.

We know that for all real values of x

$\Rightarrow -\sqrt{a^{2}+b^{2}}\leq asinx+bcosx \leq \sqrt{a^{2}+b^{2}}$

$\Rightarrow -\sqrt{a^{2}+b^{2}}\leq (y-c)\leq \sqrt{a^{2}+b^{2}}$

$\Rightarrow c-\sqrt{a^{2}+b^{2}}\leq y\leq c+\sqrt{a^{2}+b^{2}}$

$\Rightarrow c-\sqrt{a^{2}+b^{2}}\leq f(x)\leq c+\sqrt{a^{2}+b^{2}}$

Hence the range of the function $asinx+bcosx+c$ is $\mathbf{\left [ c-\sqrt{a^{2}+b^{2}}, c+ \sqrt{a^{2}+b^{2}}\right ]}$

Examples

Example1. Find the range of cosx-sinx.

Here a=-1, b=1, c=0

Hence the range of cosx-sinx= $\left [ -\sqrt{(-1)^{2}+(1)^{2}} , \sqrt{(-1)^{2}+(1)^{2}} \right ]$

= $\left [ -\sqrt{2} , \sqrt{2}\right ]$

Example 2. Find the range of -3sinx-4cosx -7

Sol. Here a= -3, b=-4 and c=-7

So range of -3sinx-4cosx-7 = $\left [ -7-\sqrt{(-4)^{2}+(-3)^{2}}, -7+\sqrt{(-4)^{2}+(-3)^{2}} \right ]$

= $\left [ -7-5, -7+5 \right ]= \left [ -12,-2 \right ]$

Also read :

Bina singh

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