how do you find the exact value of tan15 degrees?

Question

you have to use the difference formula for tangent, (60 -45), but I keep getting a different answer from the back of the book, the book's answer is 2 minus the square root of 3

Answer 1

tan 15° = tan (60° - 45°)
= (tan 60° - tan 45°)/( 1 + tan60°tan45°) [tan 60° = √3 tan 45° = 1]
= (√3 - 1)/(1 + √3*1)
= ((√3 - 1)/(1 + √3) x (√3 - 1)/(√3 - 1) (to rationalise the denominator of this fraction)
= (√3 - 1)²/2
= (3 - 2√3 + 1)/2
= (4 - 2√3)/2
= 2 - √3 ..... QED

(Similarly tan 75° = 2 + √3. Try it for the experience and pactice)

Although not the way expected an alternate way is to use the double ang1e formula:

let t = tan 15°

Then tan 30° = tan (2 x 15°)
So 1/√3 = 2t/ (1 - t²) And 1/√3 = √3/3
√3(1 - t²) = 2t
√3t² + 6t - √3 = 0
By the quadratic formula ( x = [-b ± √(b² - 4ac)]/2a ) we get
t = [-6 ± √(36 + 12)]/(2√3)
= [-6 ± √(48)]/(2√3)
= [-6 ± √(16 x 3)]/(2√3)
= [-6 ± 4√3]/(2√3)
= -3/√3 ± 2
= -√3 ± 2
= 2 - √3, -2 - √3

Now why are there 2 answers to this and which one is correct?

Correctness is easy .... tan 15° > 0 so tan 15° = 2 - √3

The other answer is not that self evident and beggars the question where did it come from because the maths is totally unabiguous so this solution is a solution to our original stated equation!

Well the answer is that this also solves tan 210° (= tan (180° + 30°)
= tan (2 x 105°) (Noting that in the third quadrant tan is also positive)
And in 2nd quadrant tan 105° = tan (180° - 75°)
= - tan 75°
= - (2 + √3)

So using the double angle results you get BOTH solutions simultaneously if somewhat hidden.

Answer 2

There is more than one way to find the answer. The book may have used the half angle formula. The answers should be the same but they look different. It is possible to show that the answers are equivalent. One way to check to make sure that your answer is the same is to punch both of them into a calculator and to make sure that the decimal approximation is the same. I teach trig and I made an extra credit question on a test about showing that the answers are equivalent.

Answer 3

tan(60-45) = (tan 60 - tan 45)/(1 + tan 60 tan 45)

tan 60 = sqrt 3; tan 45 = 1

(sqrt 3 - 1)/(1 + sqrt 3)

Times top and bottom by (1 - sqrt 3)

-3 + 2 sqrt 3 - 1 all over 1 - 3

(-4 + 2 sqrt 3)/(-2)

which is 2 - sqrt 3

Answer 4

Don't worry!
tan15 = .267949...
2 - 3^.5 = .267949...

To find those easy (like 30, 15, 45) angles you can probably just look up. The inexact decimal form is normally accepted, but if not you might want to ask your teacher.

Answer 5

tan (x/2) = sin (x/2) / cos (x/2)
= sin x / (1 + cos x)
= (1 - cos x) / sin x

I skipped some steps from my source because they would be essentially unreadable in Y!A.

tan (15°) = tan (30°/2)
= ( 1 - cos(30°) ) / sin (30°)
= ( 1 - sqrt(3)/2 ) / (1/2)
= 2 ( 1 - sqrt(3)/2 )
= 2 - sqrt(3)

Answer 6

tan15*=tan(60-45)
=tan60-tan45/1+tan60tan45
tan60=rt3 and tan45=1
substituting
tan15*=rt3-1/(1+rt3)
rationalising the denominatotr
(rt3-1)^2=3+1-2rt3/2=4-2rt3 /2
=2(2-rt3)/2=2-rt3

Answer 7

use the identity

tan(A-B)=(tanA-tanB)
/(1+tanA.tanB)...(1)

let A=60deg >>>>>tanA=sqrt3
letB=45deg>>>>>>tanB=1

substitute into (1)

tan(60-45)=(sqrt3-1)
/(1+sqrt3.1)
= (sqrt3-1)/(sqrt3+1)
=(2-sqrt3)
=tan15deg (as required

i hope that this helps

Answer 8

tan[A - B] = [tanA - tanB]/[ one million + tanA.tanB] If A = forty 5 levels tanA = one million hence tan[forty 5 - B] = [one million - tanB]/[one million + tanB] yet tan[15] = tan[forty 5 - 30] = [ one million - tan30 ]/[ one million + tan30] and because tan30 = one million/(sqrt3) tan[15] = [one million - one million/(sqrt3)]/[one million + one million/(sqrt3)] = [(sqrt3) -one million]/[(sqrt3) + one million] or in words Root 3 minus one million divided via Root 3 plus one million

Answer 9

use a calculator