The maximum value of (7 costheta + 24 sinthet | vedclass

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Question

(C) Let $y = (7 \cos	heta + 24 \sin	heta)(7 \sin	heta - 24 \cos	heta)$.We can rewrite the expression as:$y = 49 \sin	heta \cos	heta - 168 \cos^2

Vedclass · Accepted Answer

$\frac{625}{2}$

Vedclass · Answer

(C) Let $y = (7 \cos	heta + 24 \sin	heta)(7 \sin	heta - 24 \cos	heta)$.We can rewrite the expression as:$y = 49 \sin	heta \cos	heta - 168 \cos^2	heta + 168 \sin^2	heta - 576 \sin	heta \cos	heta$$y = 168(\sin^2	heta - \cos^2	heta) - 527 \sin	heta \cos	heta$$y = -168 \cos(2	heta) - \frac{527}{2} \sin(2	heta)$.The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$.Here,$a = -168$ and $b = -\frac{527}{2}$.However,using the identity $y = (7 \cos	heta + 24 \sin	heta)(7 \sin	heta - 24 \cos	heta)$:Let $7 = r \cos\phi$ and $24 = r \sin\phi$,so $r = 25$.$y = (r \cos\phi \cos	heta + r \sin\phi \sin	heta)(r \cos\phi \sin	heta - r \sin\phi \cos	heta)$$y = r^2 \cos(	heta - \phi) \sin(	heta - \phi)$$y = \frac{r^2}{2} \sin(2(	heta - \phi)) = \frac{625}{2} \sin(2(	heta - \phi))$.The maximum value is $\frac{625}{2}$.

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

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