Determine the measure of θ

khaezyline

The question is for the value of sin -√3/2 determine the measure of θ if π/2 θ ≤ 3π/2.

Does it have something to do with special triangles? If so how do I use it solve the problem? The answer is 4
π/3 but I don't know how to get there.

skeeter

MHF Helper
Unit circle ... learn it, live it, love it.

2 people

khaezyline

Is there another way to answer it like an equation or something? We haven't touched on the unit circle so I dont know how to use that.

HallsofIvy

MHF Helper
Start with an equilateral triangle with each side of length 1. The line from any vertex perpendicular to the opposite side divides that triangle into two right triangles. Further, the that line bisects the angle at the vertex and bisects the opposite side. Since all three angles in an equilateral triangle are $\pi/3$ radians (60 degrees) the angle at the bisected vertex is $\pi/6$ (30 degrees).

Further, the length of the side "adjacent" to the $\pi/6$ (30 degrees) angle is $\frac{1}{2}$ and, by the Pythagorean theorem, the side "opposite" the $\pi/3$ (60 degrees) angle is $\sqrt{1- 1/4}= \sqrt{3/4}= \frac{\sqrt{3}}{2}$ while the hypotenuse has length 1.

So $sin(\pi/3)= sin(60)= \frac{\frac{\sqrt{3}}{2}}{1}= \frac{\sqrt{3}}{2}$
$cos(\pi/3)= \cos(60)= \frac{\frac{1}{2}}{1}= \frac{1}{2}$
$tan(\pi/3)= tan(60)= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}= \sqrt{3}$.
secant, cosecant, and cotangent are the reciprocals of those.

And, of course, the complementary angle, $\pi/6$ (30 degrees) has
$cos(\pi/3)= sin(60)= \frac{\frac{\sqrt{3}}{2}}{1}= \frac{\sqrt{3}}{2}$
$sin(\pi/3)= \cos(60)= \frac{\frac{1}{2}}{1}= \frac{1}{2}$
$cot(\pi/3)= tan(60)= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}= \sqrt{3}$.

Another easy angle is $\pi/4$ radians (45 degrees). Since that is half a right angle, an isosceles right triangle with legs of length 1 has both acute angles of measure $\pi/4$ radians (45 degrees) and, by the Pythagorean theorem, hypotenuse of length
$\sqrt{2}$.

So $sin(\pi/4)= sin(45)= cos(\pi/4)= cos(45)= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$
$tan(\pi/4)= tan(45)= cot(\pi/4)= cot(45)= \frac{1}{1}= 1$.

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