# Thread: Cant solve If 𝑓(𝜃) = 2sin𝜃 − 𝑠𝑖𝑛 (𝜃/2) then find 𝑓(𝜋/3).

1. ## Cant solve If 𝑓(𝜃) = 2sin𝜃 − 𝑠𝑖𝑛 (𝜃/2) then find 𝑓(𝜋/3).

Why can't you just evaluate the terms at $\theta = \tfrac\pi3$?
This is trivial if you are allowed to use a calculator. Is the difficulty that you are asked for the exact value? For these "special angles", think of an equilateral triangle. Since it is equilateral it is also equiangular- all three angles are the same. Since the angles in any triangle add to $\displaystyle \pi$ radians, each of the angles in an equilateral triangle las measure $\displaystyle \pi/3$. Now draw a line from any vertex perpendicular to the opposite side. It is fairly easy to prove that this line also bisects the vertex angle and bisects the side. That is, it divides the equilateral triangle into two congruent right triangles in which one non-right angle is $\displaystyle \pi/3$ and the other is half of that, $\displaystyle \pi/6$. Further, one leg is half the length of the hypotenuse which is a side of the original triangle. If we call the length of the hypotenuse s, one leg has length s/2, and we will call the length of the other leg "x". Then by the Pythagorean theorem we have $\displaystyle (s/2)^2+ x^2= s^2$ so $\displaystyle x^2= s^2- s^2/4= (3/4)s^2$. Taking the square root of both sides $\displaystyle x= \frac{\sqrt{3}}{2}s$.
The result of all of that is that an right triangle with one non-right angle $\displaystyle \pi/3$ and the other [tex]\pi/6[tex] with hypotenuse s has the leg opposite the $\displaystyle \pi/6$ angle of length $\displaystyle \frac{s}{2}$ and the other, opposite the $\displaystyle \pi/3$ angle of length $\displaystyle \frac{\sqrt{3}}{2}s$. Since sine is "opposite side over hypotenuse" we have $\displaystyle sin(\pi/6)= \frac{1}{2}$ and $\displaystyle sin(\pi/3)= \frac{\sqrt{3}}{2}$.