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

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Question

(C) Let $y = (7 \cos	heta + 24 \sin	heta) 	imes (7 \sin	heta - 24 \cos	heta)$.We know that $7 \cos	heta + 24 \sin	heta = r \sin(	heta + \alpha

Vedclass · Accepted Answer

$\frac{625}{2}$

Vedclass · Answer

(C) Let $y = (7 \cos	heta + 24 \sin	heta) 	imes (7 \sin	heta - 24 \cos	heta)$.We know that $7 \cos	heta + 24 \sin	heta = r \sin(	heta + \alpha)$,where $r = \sqrt{7^2 + 24^2} = 25$ and $	an \alpha = \frac{7}{24}$.Similarly,$7 \sin	heta - 24 \cos	heta = r \sin(	heta - \beta)$,where $	an \beta = \frac{24}{7}$.Alternatively,let $7 \cos	heta + 24 \sin	heta = 25 \sin(	heta + \alpha)$ and $7 \sin	heta - 24 \cos	heta = -25 \cos(	heta + \alpha)$.Then $y = 25 \sin(	heta + \alpha) 	imes (-25 \cos(	heta + \alpha)) = -625 \sin(	heta + \alpha) \cos(	heta + \alpha)$.Using the identity $\sin(2x) = 2 \sin x \cos x$,we get $y = -\frac{625}{2} \sin(2(	heta + \alpha))$.The maximum value of $\sin(2(	heta + \alpha))$ is $1$ and the minimum is $-1$.Since we are looking for the maximum value of the expression,we consider the magnitude or the range. However,the expression simplifies to $-\frac{625}{2} \sin(2(	heta + \alpha))$.The maximum value of this expression is $|-\frac{625}{2}| = \frac{625}{2}$.

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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