March 26, 2023

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

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 $sin(\alpha&space;+\beta&space;)=sin\alpha&space;cos\beta&space;+cos\alpha&space;sin\beta$   we have  $\large&space;asinx+bcosx$ . So we would  like to find an angle $\beta$ such that  $cos\beta&space;=a$ and $sin\beta&space;=b$, for then we could write    $asinx+bcosx=cos\beta&space;sinx+sin\beta&space;cosx=sin(x+\beta&space;)$

Since $sin\beta$ and $cos\beta$ must be between 1 and 1, and $a$ and $b$ may not be in that range. Moreover, we know that $cos^{2&space;}\beta&space;+sin^{2}\beta$ 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  $cos\beta&space;=A$  and $sin\beta&space;=B$ and $0\leq&space;\beta&space;<&space;2\pi$. Then $asinx+bcosx&space;=\sqrt{a^{2}+b^{2}}(Asinx+Bcosx)$

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

$=\sqrt{a^{2}+b^{2}}sin(x+\beta&space;)$
so $y=\sqrt{a^{2}+b^{2}}sin(x+\beta&space;)$.
Since $-1\leq&space;sin(x+\beta&space;)\leq&space;1$, this implies
$=-\sqrt{a^{2}+b^{2}}\leq&space;\sqrt{a^{2}+b^{2}}sin(x+\beta&space;)\leq&space;\sqrt{a^{2}+b^{2}}$
$\Rightarrow&space;-\sqrt{a^{2}+b^{2}}\leq&space;y\leq&space;\sqrt{a^{2}+b^{2}}$.

## 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

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

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

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

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

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

### Examples

Example1. Find the range of  cosx-sinx.

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

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

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

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

Sol.  Here a= -3, b=-4 and c=-7

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

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