using half identities to find exact value of each trigonometric expression a) $\cos{({105}^{°})}$
b) $\sin{(\frac{3\pi}{8})}$
c) $\cot{({67.5}^{°})}$
please do explain how you are able to get the answer as I'm still confused about this topic... Thank you
 A: HINT. 
For the first one you just need the compound angle identity, i.e. $\cos(60 +45)$
For the second one, use $\sin^2x=\frac 12(1-\cos2x)$
For the third one, try $\csc2x+\cot2x=\cot x$
A: Notice, $$\cos 105^\circ=\cos(90^\circ+15^\circ)=-\sin 15^\circ=-\sqrt{\frac{1-\cos 30^\circ}{2}}=-\sqrt{\frac{1-\frac{\sqrt 3}{2}}{2}}=-\sqrt{\frac{2-\sqrt 3}{4}}$$ $$=-\sqrt{\frac{4-2\sqrt 3}{8}}= -\frac{\sqrt 3-1}{2\sqrt 2}=\frac{1-\sqrt 3}{2\sqrt 2}$$ $$\implies \sin\frac{3\pi}{8}=\sin 67.5^\circ=\sin(90^\circ-22.5^\circ)=\cos 22.5^\circ=-\sqrt{\frac{1+\cos 45^\circ}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt2}}{2}}=\sqrt{\frac{\sqrt 2+1}{2\sqrt 2}}=\sqrt{\frac{2+\sqrt 2}{4}}=\frac{\sqrt{2+\sqrt 2}}{2}$$ 
$$\implies \cot 67.5^\circ=\sqrt{\csc^267.5-1}=\sqrt{\frac{1}{\sin^2 67.5^\circ}-1}=\sqrt{\left(\frac{2}{\sqrt{2+\sqrt 2}}\right)^2-1}=\sqrt{\frac{2-\sqrt2}{2+\sqrt 2}}=\sqrt{\frac{(2-\sqrt2)^2}{4-2}}=\frac{2-\sqrt 2}{\sqrt2}=\sqrt 2-1$$
Thus, we have all the values as follows $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cos 105^\circ=\frac{1-\sqrt3}{2\sqrt2}}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\sin \frac{3\pi}{8}=\frac{\sqrt{2+\sqrt2}}{2}}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cot 67.5^\circ=\sqrt 2-1}}$$
A: $(1)$
$$\cot A=\dfrac{\cos A}{\sin A}=\dfrac{2\cos^2A}{2\cos A\sin A}=\dfrac{1+\cos2A}{\sin2A}$$
or $$\cot A=\cdots=\dfrac{\sin2A}{1-\cos2A}$$

$(2)$
$$\cos2B=2\cos^2B-1=1-2\sin^2B$$
$$\cos^2B=\dfrac{1+\cos2B}2,\cos B=\pm\sqrt{\dfrac{1+\cos2B}2}$$
Now use All Sin Tan Cos Rule 
