The value of a cos theta + b sin theta lies b | vedclass

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(D) Let $f(	heta) = a \cos 	heta + b \sin 	heta$.We can write this as $f(	heta) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos 	heta +

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$-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$

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(D) Let $f(	heta) = a \cos 	heta + b \sin 	heta$.We can write this as $f(	heta) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos 	heta + \frac{b}{\sqrt{a^2 + b^2}} \sin 	heta ight)$.Let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$.Then $f(	heta) = \sqrt{a^2 + b^2} (\cos 	heta \cos \phi + \sin 	heta \sin \phi) = \sqrt{a^2 + b^2} \cos(	heta - \phi)$.Since the range of $\cos(	heta - \phi)$ is $[-1, 1]$,the range of $f(	heta)$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Therefore,the value of $a \cos 	heta + b \sin 	heta$ lies between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.

The value of $a \cos \theta + b \sin \theta$ lies between

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