Let the product of omega_1=(8+i) sin theta+(7 | vedclass

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(B) Given $\omega_1 = (8 \sin 	heta + 7 \cos 	heta) + i(\sin 	heta + 4 \cos 	heta)$ and $\omega_2 = (\sin 	heta + 4 \cos 	heta) + i(8 \sin 	het

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(B) Given $\omega_1 = (8 \sin 	heta + 7 \cos 	heta) + i(\sin 	heta + 4 \cos 	heta)$ and $\omega_2 = (\sin 	heta + 4 \cos 	heta) + i(8 \sin 	heta + 7 \cos 	heta)$.Let $x = 8 \sin 	heta + 7 \cos 	heta$ and $y = \sin 	heta + 4 \cos 	heta$.Then $\omega_1 = x + iy$ and $\omega_2 = y + ix$.The product $\omega_1 \omega_2 = (x + iy)(y + ix) = xy + ix^2 + iy^2 + i^2yx = xy + i(x^2 + y^2) - xy = i(x^2 + y^2)$.Thus,$\alpha = 0$ and $\beta = x^2 + y^2$.$\beta = (8 \sin 	heta + 7 \cos 	heta)^2 + (\sin 	heta + 4 \cos 	heta)^2 = (64 \sin^2 	heta + 49 \cos^2 	heta + 112 \sin 	heta \cos 	heta) + (\sin^2 	heta + 16 \cos^2 	heta + 8 \sin 	heta \cos 	heta) = 65 \sin^2 	heta + 65 \cos^2 	heta + 120 \sin 	heta \cos 	heta = 65 + 60 \sin(2 	heta)$.Since $\alpha + \beta = 0 + 65 + 60 \sin(2 	heta) = 65 + 60 \sin(2 	heta)$,the maximum value $p = 65 + 60 = 125$ and the minimum value $q = 65 - 60 = 5$.Therefore,$p + q = 125 + 5 = 130$.

Let the product of $\omega_1=(8+i) \sin \theta+(7+4 i) \cos \theta$ and $\omega_2=(1+8 i ) \sin \theta+(4+7 i ) \cos \theta$ be $\alpha+ i \beta$,where $i =\sqrt{-1}$. Let $p$ and $q$ be the maximum and the minimum values of $\alpha+\beta$ respectively. Then the value of $p+q$ is equal to

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