186k views
3 votes
Find the exact value of cot(15degrees) using half angle formulas?

2 Answers

2 votes

Answer:

C

Explanation:

C on edg

User Jamie Babineau
by
8.6k points
4 votes

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


\cos^2\frac x2=\frac{1+\cos x}2


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


\implies\cot^2\frac x2=(\cos^2\frac x2)/(\sin^2\frac x2)=(1+\cos x)/(1-\cos x)

So we have


\cot^215^\circ=(1+\cos30^\circ)/(1-\cos30^\circ)=\frac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}


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


\implies\cot15^\circ=√(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)

User AlexPad
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.