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IceJOKER [234]
3 years ago
6

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

Mathematics
2 answers:
liubo4ka [24]3 years ago
8 0

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)

lesantik [10]3 years ago
6 0

Answer:

C

Step-by-step explanation:

C on edg

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