Answer:
1) cos 105°
2) tan 67.5°
3) sin 67.5°
4) tan 165°
Step-by-step explanation:
From half angle identity
[tex]sin\frac{\theta }{2}= \sqrt{\frac{1-cos\theta }{2}}[/tex]
For sin 67.5 = sin (135/2)
Here 67.5° lies in Ist quadrant therefore
[tex]sin\frac{\theta }{2}= \sqrt{\frac{1-cos\theta }{2}}[/tex]
[tex]=\sqrt{\frac{(1-cos135)}{2}}=\sqrt{\frac{1+cos45}{2}}[/tex]
[tex]=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}[/tex]
[tex]=\sqrt{\frac{(\sqrt{2}+1)(\sqrt{2})}{2\sqrt{2}\times \sqrt{2}}}[/tex]
[tex]=\sqrt{\frac{\sqrt{2}+2}{2}}=\frac{\sqrt{\sqrt{2}+2}}{2}[/tex]
For cos 105 = cos (210/2)
From half angle identity
[tex]cos(\theta/2) =\pm \sqrt{\frac{1+cos\theta }{2}}[/tex]
Since cos 105 lies in second quadrant
Therefore [tex]cos(\theta/2) =-\sqrt{\frac{1+cos\theta }{2}}[/tex]
[tex]cos(\210/2) =-\sqrt{\frac{1+cos210}{2}}[/tex]
[tex]=-\sqrt{\frac{1-cos60}{2}}=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}[/tex]
[tex]=-\sqrt{\frac{2-\sqrt{3}}{4}}=-\frac{\sqrt{2-\sqrt{3}}}{2}[/tex]
For tan165 = tan (330/2)
From half angle identity
[tex]tan(\theta/2) =\frac{sin\theta }{1+cos\theta }[/tex]
[tex]=\frac{sin330}{1+cos330}=-\frac{sin30}{1+cos30}[/tex]
[tex]=-\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}=-\frac{1}{2+\sqrt{3}}[/tex]
[tex]=-\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=-\frac{2-\sqrt{3}}{4-3}=-(2-\sqrt{3})=\sqrt{3}-2[/tex]
For tan 67.5 = tan (135/2)
From half angle identity
[tex]tan(\theta/2) =\frac{sin\theta }{1+cos\theta }[/tex]
[tex]=\frac{sin135}{1+cos135}=\frac{sin45}{1+cos45}[/tex]
[tex]=\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}[/tex]
[tex]=\frac{1}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}[/tex]
[tex]=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1[/tex]
