If sin theta + cos theta = frac{1}{2},then 16 | vedclass

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(C) Given $\sin 	heta + \cos 	heta = \frac{1}{2}$.Squaring both sides: $(\sin 	heta + \cos 	heta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + \sin(2	he

Vedclass · Accepted Answer

$-23$

Vedclass · Answer

(C) Given $\sin 	heta + \cos 	heta = \frac{1}{2}$.Squaring both sides: $(\sin 	heta + \cos 	heta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + \sin(2	heta) = \frac{1}{4}$ $\Rightarrow \sin(2	heta) = -\frac{3}{4}$.Then $\cos^2(2	heta) = 1 - \sin^2(2	heta) = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.We need to evaluate $16(\sin(2	heta) + \cos(4	heta) + \sin(6	heta))$.Using sum-to-product formula: $\sin(2	heta) + \sin(6	heta) = 2\sin(4	heta)\cos(2	heta)$.So the expression is $16(2\sin(4	heta)\cos(2	heta) + \cos(4	heta))$.Since $\sin(4	heta) = 2\sin(2	heta)\cos(2	heta)$ and $\cos(4	heta) = 2\cos^2(2	heta) - 1$:$16(2(2\sin(2	heta)\cos(2	heta))\cos(2	heta) + 2\cos^2(2	heta) - 1) = 16(4\sin(2	heta)\cos^2(2	heta) + 2\cos^2(2	heta) - 1)$.Substituting values: $16(4(-\frac{3}{4})(\frac{7}{16}) + 2(\frac{7}{16}) - 1) = 16(-\frac{21}{16} + \frac{14}{16} - 1) = 16(-\frac{7}{16} - 1) = -7 - 16 = -23$.

If $\sin \theta + \cos \theta = \frac{1}{2}$,then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to:

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