Question

Find the exact value

Answer 1

Answer:

A

Step-by-step explanation:

$cos~ \theta=\sqrt{1-sin^2 \theta} =\sqrt{1-(\frac{1}{2})^2 } =\frac{\sqrt{3} }{2} \\1+cos~\theta=2cos^2\frac{\theta}{2} \\1+\frac{\sqrt{3}}{2} =2~cos^2 \frac{\theta}{2} \\cos^2\frac{\theta}{2} =\frac{2+\sqrt{3}}{2 \times 2} \\cos \frac{\theta}{2} =\frac{\sqrt{2+\sqrt{3}}}{2} \\1-cos \theta=2 ~sin^2\frac{\theta}{2} \\1-\frac{\sqrt{3}}{2}=2~sin^2 \frac{\theta}{2} \\sin^2 \frac{\theta}{2} =\frac{2-\sqrt{3} }{2 \times 2} \\sin \frac{\theta}{2}=\frac{\sqrt{2-\sqrt{3} } }{2}$

as 0≤θ≤90

so θ/2 is also in 0≤θ≤90

hence sin θ/2 and cos θ/2 are positive.