Tan(15/2)=√6-√3+√2-2
Answers
Answer: I am assuming you want to prove the (true) statement:
tan(7.5°) = √6 - √3 + √2 - 2;
Step-by-step explanation: in order to do this, we need to use the half-angle formula for tangent:
tan(θ/2) = sinθ/(1 + cosθ) = (1 - cosθ)/sinθ.
To limit the number of denominator rationalizations needed, we should use:
tan(θ/2) = (1 - cosθ)/sinθ.
By letting θ = 15°, we get:
tan(7.5°) = [1 - cos(15°)]/sin(15°).
To find cos(15°) and sin(15°), we could use the sine and cosine half-angle formulas, but we have would radicals nested inside radicals, which would be hard to simplify; instead, we can use the sine and cosine subtraction formulas.
We have:
(a) cos(15°) = cos(45° - 30°), by writing 15° as a difference of two special angles
= cos(45°)cos(30°) + sin(45°)sin(30°), by the cosine subtraction formula
= (√2/2)(√3/2) + (√2/2)(1/2)
= (√6 + √2)/4
(b) sin(15°) = sin(45° - 30°), by doing the same as (a)
= sin(45°)cos(30°) - cos(45°)sin(30°), by the sine subtraction formula
= (√2/2)(√3/2) - (√2/2)(1/2)
= (√6 - √2)/4.
Thus:
tan(7.5°) = [1 - cos(15°)]/sin(15°)
= [1 - (√6 + √2)/4]/[(√6 - √2)/4], from above
= (4 - √6 - √2)/(√6 - √2), by multiplying the numerator and denominator by 4
= [(4 - √6 - √2)(√6 + √2)]/(6 - 2), by rationalizing the denominator
= (4√6 - 6 - 2√3 + 4√2 - 2√3 - 2)/4
= (4√6 - 4√3 + 4√2 - 8)/2
= √6 - √3 + √2 - 2, as required.
I hope this helps!
And yes sorry for the long answer