April 13, 2024

How to find the range of asinx+bcosx+c



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


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

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 }\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 ]}


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 ]


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