Question

Find the exact value of cot(15degrees) using half angle formulas?

Answer 1

Double angle (or half angle, depending how you look at it) identities:

$\cos^2\dfrac x2=\dfrac{1+\cos x}2$

$\sin^2\dfrac x2=\dfrac{1-\cos x}2$

$\implies\cot^2\dfrac x2=\dfrac{\cos^2\frac x2}{\sin^2\frac x2}=\dfrac{1+\cos x}{1-\cos x}$

So we have

$\cot^215^\circ=\dfrac{1+\cos30^\circ}{1-\cos30^\circ}=\dfrac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}$

$\implies\cot^215^\circ=7+4\sqrt3$

$\implies\cot15^\circ=\sqrt{7+4\sqrt3}=2+\sqrt3$

Note that when taking the square root, we should take into account that that could yield two possible solutions, but we know $\cos15^\circ>0$ and $\sin15^\circ>0$, so it's also the case that $\cot15^\circ>0$.

Also, the reason we have equality in the last step can be explained like so:

$7+4\sqrt3=4+4\sqrt3+3=4+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2$

(not unlike the process used to complete the square)

Answer 2

Answer:

C

Step-by-step explanation:

C on edg