Answer:
Step-by-step explanation:
sin θ=1/4,tan θ>0
so θ lies in first quadrant.
cos θ=√(1-sin²θ)=√(1-(1/4)²)=√(1-1/16)=√(15/16)=√15/4
[tex]2cos^2 \frac{\theta}{2} =1+cos \theta[/tex]
[tex]2~cos^2\frac{\theta}{2} =1+\frac{\sqrt{15} }{4} =\frac{4+\sqrt{15}}{4} \\cos~\frac{\theta}{2} =\sqrt{\frac{4+\sqrt{15} }{4 \times 2} } \\cos\frac{\theta}{2} =\frac{\sqrt{8+2\sqrt{15}} }{4}[/tex]
