Determine the measure of θ

Dec 2017
4
0
Canada
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
Jun 2008
16,217
6,765
North Texas
Unit circle ... learn it, live it, love it.

 
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Dec 2017
4
0
Canada
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
Apr 2005
20,249
7,909
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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