Tan 15 degrees ÷ 2- 2tan squared 15 degrees

Answer 1

Hi there Karekezi,

The problem is:

tan(15°)/2 - 2tan² (15°)

Now, this can be solved with a calculator, but I feel it is very useful to work through it logically. In this problem, it is important to recognize that we can use the angle addition formula and our knowledge of the side ratios of special triangles to solve this equation.

The angle addition and subtraction formulas for a tangent:

tan(Θ + β) = (tan(Θ) + tan(β)) / (1 − tan(Θ)tan(β))

and since we know tan(-β) = -tan(β) we can derive the subtraction formula by multiplying -1 times tan(β) in the equation

tan(Θ - β) = (tan(Θ) - tan(β)) / (1 + tan(Θ)tan(β))

I feel like the intuition for how this formula is derived is important, but I won't go into too much right here.

Now, if you remember your special triangle cases:

For triangles with the angles 45°, 45°, and 90°,

Hypotenuse = √2

Other 2 sides = 1

and

For triangles with angles 30°, 60°, and 90°,

Hypotenuse = 1

The side opposite the 30° angle = 1/2

The side opposite the 60° angle = (√3)/2

The problem asks for the tangent of 15° but that is not something that everyone knows by heart. One way we can figure it out is by using the fact that the tan(45°-30°) gives us the tangent of 15°. Now we know that the tangent of 45° is 1 and the tangent of 30° is 1/√3 and we can plug that into the angle subtraction equation. Now we have

tan(45° - 30°) = (tan(45°) - tan(30°)) / (1 + tan(45°)tan(30°))

substitute for known values

tan(45° - 30°) = (1 - 1/√3) / (1 + 1/√3)

simplify fractions by multiply by √3/√3

tan(45° - 30°) = (√3 - 1) / (√3 + 1)

If we want to simplify further we can multiply by the conjugate base to remove the root from the denominator using our knowledge of the difference of squares. (Ex: x⁴ - 9 = (x² + 3)(x² - 3))

tan(45° - 30°) = (√3 - 1) / (√3 + 1) * (√3 - 1) / (√3 - 1)

simplify

tan(45° - 30°) = (3 - 2√3 + 1) / (3 - 1)

tan(45° - 30°) = (4 - 2√3) / 2

tan(45° - 30°) = 2 - √3

And drumroll....

tan(15°) = 2 - √3

Now that we finally know what the tangent of 15 degrees is we can plug it into the original equation.

tan(15°)/2 - 2tan² (15°) = (2 - √3) / 2 - 2(2 - √3)²

simplify

= 1 - (√3)/2 - 8 + 8√3 - 6

= - (√3)/2 + 8√3 - 13

= (- √3 + 16√3) / 2 - 13

And second drumroll....

Answer = (15/2)√3 - 13

Answer 2

sin 30° = 1/2

cos 30°= √3/2

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

Then, tan 15° /(2 - 2 tan²15° ) =

(2 - √3)/(2 - 2 * (2 - √3)²) =

(2 - √3)/(2 - 2 * (7 - 4√3)) =

(2 - √3)/(2 - 14 + 8√3) =

(2 - √3)/(-12 + 8√3) =

(2 - √3)/4/(2√3 - 3) = (multiplying top and bottom by 2√3 + 3)

√3/12