For angles [tex]\theta[/tex] between 0º and 90º, we know [tex]\sin\theta[/tex] and [tex]\cos\theta[/tex] are both positive.
Then [tex]\frac\theta2[/tex] falls between 0º and 45º, so both [tex]\sin\frac\theta2[/tex] and [tex]\cos\frac\theta2[/tex] are also positive.
Recall the double angle identities,
[tex]\sin^2x=\dfrac{1-\cos2x}2\implies\sin^2\dfrac\theta2=\dfrac{1-\cos\theta}2[/tex]
[tex]\cos^2x=\dfrac{1+\cos2x}2\implies\cos^2\dfrac\theta2=\dfrac{1+\cos\theta}2[/tex]
We know sine and cosine should be positive, so taking the square root of both sides gives us
[tex]\sin\dfrac\theta2=\sqrt{\dfrac{1-\cos\theta}2}[/tex]
[tex]\cos\dfrac\theta2=\sqrt{\dfrac{1+\cos\theta}2}[/tex]
Also recall the Pythagorean identity,
[tex]\cos^2x=1-\sin^2x\implies\cos\theta=\sqrt{1-\sin^2\theta}[/tex]
Then
[tex]\cos\theta=\sqrt{1-\left(\dfrac27\right)^2}=\dfrac{3\sqrt5}7[/tex]
and from here we get
[tex]\sin\dfrac\theta2=\sqrt{\dfrac{7-3\sqrt5}{14}}[/tex]
[tex]\cos\dfrac\theta2=\sqrt{\dfrac{7+3\sqrt5}{14}}[/tex]
