One more trig. problem

Given that tan 75 = 2 + root 3, find in the form m + n root 3, where m and n are integers, the value of tan 15.

I can't get this one. There's another one which is tan 105 but I got that straight away, its - 2 - root 3. Can anyone tell me how to work out this problem please?

Reply 1

Juwel

mik1a

A start would be that a right angled triangle with another angle of 75 has third angle of 15, so begin by drawing said triangle...

Reply 2

john !!

so tan 75 would be the opposite side over the adjacent side. so i convert 2+root3 into a fraction... (3+2root3)/(root3) therefore tan 15 must be (root3) / (3+2root3) ?

I know that's wrong ...

Reply 3

JamesF

Tan 15 = tan(75 - 60)
tan(75-60) = (tan 75 - tan 60)/(1 + tan75tan60)

tan60 = tan(30+30) = tan 30 + tan 30)/(1 - tan^2 30)
You should know that tan 30 = sqrt(1/3)

So tan 60 = 2sqrt(1/3)/(1-1/3) = 3sqrt(1/3) = sqrt(3).

tan 15 = (2 + sqrt 3 - sqrt3)/(1+3+2sqrt(3))
tan 15 = 2/(4+2sqrt3) = 1/(2 + sqrt3)

Rationalise the denominator to get
tan 15 = (2-sqrt3)/(4 - 3) = 2-sqrt3

Reply 4

JamesF

One quick trick that i like is this.
tan(90 - x) = 1/(tan x)
This greatly simplifies your problem since 15 = 90-75

tan 90-75 = 1/tan75 = 1/2+sqrt3 = 2-sqrt3.

Reply 5

Juwel

JamesF

One quick trick that i like is this.
tan(90 - x) = 1/(tan x)
This greatly simplifies your problem since 15 = 90-75

tan 90-75 = 1/tan75 = 1/2+sqrt3 = 2-sqrt3.

Nice trick, how does it apply to the other trigonometric functions?