need help with part b of this question: prove that tan(pi/12)= sqrt(7 - 4sqrt3)

[hndalama]

a.) Prove the identity (1-tan²x) / (1 + tan²x) = cos2x

b.) Hence, prove that tan(pi/12)= sqrt(7 - 4sqrt3)

I have managed to proved the identity but I don't know what to do for part b.

 

[Steven G]

a.) Prove the identity (1-tan²x) / (1 + tan²x) = cos2x

b.) Hence, prove that tan(pi/12)= sqrt(7 - 4sqrt3)

I have managed to proved the identity but I don't know what to do for part b.

use tan(pi/12)= sqrt(7 - 4sqrt3) in the formula.

Compute [1-(sqrt(7 - 4sqrt3)^2)]/[1+(sqrt(7 - 4sqrt3)^2)] and see if it equals cos(2*pi/12). Then what can you say?

 

[Ishuda]

a.) Prove the identity (1-tan²x) / (1 + tan²x) = cos2x

b.) Hence, prove that tan(pi/12)= sqrt(7 - 4sqrt3)

I have managed to proved the identity but I don't know what to do for part b.

Suppose you have
\(\displaystyle \dfrac{1-a}{1+a}\, =\, b\)
What is a in terms of b? What is cos(2x) when x=\(\displaystyle \dfrac{\pi}{12}\)?

 

[hndalama]

Thank you guys, I know how to solve it now

Suppose you have
\(\displaystyle \dfrac{1-a}{1+a}\, =\, b\)
What is a in terms of b?

@Ishuda My understanding is that a cannot be expressed in terms of b. How would you do that?

 

[Deleted member 4993]

Thank you guys, I know how to solve it now



@Ishuda My understanding is that a cannot be expressed in terms of b. How would you do that?

Multiply "both sides" (of the equation) by (1 + a). What do you get?

 

[hndalama]

Multiply "both sides" (of the equation) by (1 + a). What do you get?

Oh, It was simpler than i thought.

1- a = b + ba
1 =b + ba + a
1 - b = a (b+1)
(1 - b)/(1 + b) = a

thank you

 

[Deleted member 4993]

Oh, It was simpler than i thought.

That's the magic of working with pencil & paper - instead of staring at the screen!!