Question

Help I can't find the answer

Answer 1

For angles $\theta$ between 0º and 90º, we know $\sin\theta$ and $\cos\theta$ are both positive.

Then $\frac\theta2$ falls between 0º and 45º, so both $\sin\frac\theta2$ and $\cos\frac\theta2$ are also positive.

Recall the double angle identities,

$\sin^2x=\dfrac{1-\cos2x}2\implies\sin^2\dfrac\theta2=\dfrac{1-\cos\theta}2$

$\cos^2x=\dfrac{1+\cos2x}2\implies\cos^2\dfrac\theta2=\dfrac{1+\cos\theta}2$

We know sine and cosine should be positive, so taking the square root of both sides gives us

$\sin\dfrac\theta2=\sqrt{\dfrac{1-\cos\theta}2}$

$\cos\dfrac\theta2=\sqrt{\dfrac{1+\cos\theta}2}$

Also recall the Pythagorean identity,

$\cos^2x=1-\sin^2x\implies\cos\theta=\sqrt{1-\sin^2\theta}$

Then

$\cos\theta=\sqrt{1-\left(\dfrac27\right)^2}=\dfrac{3\sqrt5}7$

and from here we get

$\sin\dfrac\theta2=\sqrt{\dfrac{7-3\sqrt5}{14}}$

$\cos\dfrac\theta2=\sqrt{\dfrac{7+3\sqrt5}{14}}$