The extreme values of 1 + 4 sin theta + 3 cos | vedclass

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Question

(C) Given expression is $E = 1 + 4 \sin 	heta + 3 \cos 	heta$.We know that the expression $a \sin 	heta + b \cos 	heta$ lies in the interval $[-\s

Vedclass · Accepted Answer

Both $6$ and $-4$

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(C) Given expression is $E = 1 + 4 \sin 	heta + 3 \cos 	heta$.We know that the expression $a \sin 	heta + b \cos 	heta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 4$ and $b = 3$.So,$\sqrt{a^2 + b^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.Thus,the range of $4 \sin 	heta + 3 \cos 	heta$ is $[-5, 5]$.Now,adding $1$ to the entire range,we get the range of $E$ as $[1 - 5, 1 + 5]$,which is $[-4, 6]$.Therefore,the extreme values are $-4$ and $6$.

The extreme values of $1 + 4 \sin \theta + 3 \cos \theta$ are:

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