(h) $$ \left(\cos \frac{ heta}{2}+\sin \frac{ heta}{2} ight)^{2}=1+\sin heta $$

Question

(h) $$ \left(\cos \frac{	heta}{2}+\sin \frac{	heta}{2}ight)^{2}=1+\sin 	heta $$

UpStudy AI · Accepted Answer

We start with the left-hand side:

$$
\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2} = \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2} + \sin^2 \frac{\theta}{2}.
$$

Using the Pythagorean identity

$$
\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} = 1,
$$

the expression becomes

$$
1 + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}.
$$

Next, recall the sine double-angle formula:

$$
\sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}.
$$

Substituting this into the expression gives

$$
1 + \sin \theta.
$$

Thus, we have shown that

$$
\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2} = 1+\sin \theta.
$$
$$ \left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2} = 1+\sin \theta $$

(h)

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