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

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(D) Let $E = 1 + 4 \sin 	heta + 3 \cos 	heta$. We know that the expression $a \sin 	heta + b \cos 	heta$ lies in the interval $[-\sqrt{a^2 + b^2},

Vedclass · Accepted Answer

Both $-4$ and $6$

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(D) Let $E = 1 + 4 \sin 	heta + 3 \cos 	heta$. 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 expression $4 \sin 	heta + 3 \cos 	heta$ ranges from $-5$ to $5$. Adding $1$ to the entire range,we get $1 + (-5) \leq E \leq 1 + 5$. Therefore,$-4 \leq E \leq 6$. The extreme values are $-4$ and $6$.

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

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