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

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Question

(D) Given expression: $E = \sin 	heta + \cos 	heta + \sin 2	heta$.We know that $\sin 2	heta = 2 \sin 	heta \cos 	heta$.Let $t = \sin 	heta + \c

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$	an \frac{3\pi}{8}$

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(D) Given expression: $E = \sin 	heta + \cos 	heta + \sin 2	heta$.We know that $\sin 2	heta = 2 \sin 	heta \cos 	heta$.Let $t = \sin 	heta + \cos 	heta$. Squaring both sides,we get $t^2 = \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 1 + \sin 2	heta$.Thus,$\sin 2	heta = t^2 - 1$.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: $E = t + t^2 - 1 = t^2 + t - 1$.To find the maximum value of $E$ on the interval $[-\sqrt{2}, \sqrt{2}]$,we evaluate the quadratic $f(t) = t^2 + t - 1$ at the boundaries.At $t = \sqrt{2}$,$E = (\sqrt{2})^2 + \sqrt{2} - 1 = 2 + \sqrt{2} - 1 = 1 + \sqrt{2}$.At $t = -\sqrt{2}$,$E = (-\sqrt{2})^2 - \sqrt{2} - 1 = 2 - \sqrt{2} - 1 = 1 - \sqrt{2}$.Since $1 + \sqrt{2} \approx 2.414$ and $	an(\frac{3\pi}{8}) = 	an(67.5^\circ) = \sqrt{2} + 1$,the maximum value is $1 + \sqrt{2}$,which corresponds to $	an \frac{3\pi}{8}$.

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

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