The maximum value of the expression E = sin t | vedclass

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Question

(B) Let $E = \sin 	heta + \cos 	heta + \sin 2	heta$.Let $t = \sin 	heta + \cos 	heta$. Then $t^2 = \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta

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$1 + \sqrt{2}$

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(B) Let $E = \sin 	heta + \cos 	heta + \sin 2	heta$.Let $t = \sin 	heta + \cos 	heta$. Then $t^2 = \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 1 + \sin 2	heta$,so $\sin 2	heta = t^2 - 1$.The range of $t = \sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \frac{\pi}{4})$ is $[-\sqrt{2}, \sqrt{2}]$.Substituting $t$ into the expression,we get $E = t + t^2 - 1 = t^2 + t - 1$.This is a quadratic in $t$ opening upwards. The maximum value on the interval $[-\sqrt{2}, \sqrt{2}]$ occurs at the boundary $t = \sqrt{2}$.$E_{\max} = (\sqrt{2})^2 + \sqrt{2} - 1 = 2 + \sqrt{2} - 1 = 1 + \sqrt{2}$.Note that $1 + \sqrt{2} = 	an(\frac{3\pi}{8})$.

The maximum value of the expression $E = \sin \theta + \cos \theta + \sin 2\theta$ is-

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