New to trigonometry

**davidman** · January 23rd 2010, 07:20 PM

Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

There are two values that fulfil $sin\theta=\frac{\sqrt{3}}{2}$

they are $\theta=\:\:,$

second part of the question:

for $sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,

if $\theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then

$\theta_3-\theta_1=\theta_4-\theta_2=$

and $\theta_2-\theta_1=\theta_4-\theta_3=$

thanks for any help.

**Prove It** · January 23rd 2010, 07:32 PM

Originally Posted by davidman

Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

There are two values that fulfil $sin\theta=\frac{\sqrt{3}}{2}$

they are $\theta=\:\:,$

second part of the question:

for $sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,

if $\theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then

$\theta_3-\theta_1=\theta_4-\theta_2=$

and $\theta_2-\theta_1=\theta_4-\theta_3=$

thanks for any help.

First, draw yourself an equilateral triangle of side length $2$ units.

Then draw a straight line perpendicular to the base and going through the "top" angle.

You will now have two $30^\circ, 60^\circ, 90^\circ$ triangles.

Their dimensions are $1, \sqrt{3}, 2$ (you can check this using Pythagoras).

Remember that $\sin{\theta} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}$ .

Which angle (out of $30^\circ$ and $60^\circ$ ) has the Opposite side with length $\sqrt{3}$ units and the Hypotenuse as $2$ units?

**VonNemo19** · January 23rd 2010, 07:34 PM

Originally Posted by davidman

Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

There are two values that fulfil $sin\theta=\frac{\sqrt{3}}{2}$

they are $\theta=\:\:,$

second part of the question:

for $sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,

if $\theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then

$\theta_3-\theta_1=\theta_4-\theta_2=$

and $\theta_2-\theta_1=\theta_4-\theta_3=$

thanks for any help.

I think that you mean to say that $\sin\theta=\frac{\sqrt{3}}{2}$ has two solutions for $0\leq\theta\leq2\pi$ ? And you wish to find the solutions?

Well, $\sqrt3/2$ is given by evaluating $\sin\theta$ at what is commonly referred to as one of the "special angles". Unfortunately, I can't draw too well on this forum, but here's something for you to look at:

Math Concepts Explained: Special Angles in Trigonometry

**bigwave** · January 23rd 2010, 07:35 PM

Originally Posted by davidman

Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

There are two values that fulfil $sin\theta=\frac{\sqrt{3}}{2}$

they are $\theta=\:\:,$

second part of the question:

for $sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,

if $\theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then

$\theta_3-\theta_1=\theta_4-\theta_2=$

and $\theta_2-\theta_1=\theta_4-\theta_3=$

thanks for any help.

$sin\theta=\frac{\sqrt{3}}{2}$

$\frac{\pi}{3},\frac{2\pi}{3}$

these are quadrants 1 and 2 respectfully

**davidman** · January 23rd 2010, 08:45 PM

Originally Posted by Prove It

Which angle (out of $30^\circ$ and $60^\circ$ ) has the Opposite side with length $\sqrt{3}$ units and the Hypotenuse as $2$ units?

$sin60=\frac{\sqrt{3}}{2}\:\:sin30=\frac{1}{2}\:\:s in90=1$

Thanks!

Originally Posted by VonNemo19

I think that you mean to say that $\sin\theta=\frac{\sqrt{3}}{2}$ has two solutions for $0\leq\theta\leq2\pi$ ? And you wish to find the solutions?

Exactly that, yes.

Originally Posted by bigwave

$sin\theta=\frac{\sqrt{3}}{2}$

$\frac{\pi}{3},\frac{2\pi}{3}$

these are quadrants 1 and 2 respectfully

this brings me to my next question... how do you get two answers when I only get one from evaluating triangles with "special angles"? And how to go from degrees to radian... ?

**bigwave** · January 23rd 2010, 09:00 PM

you may be looking at a triangle in just one quadrant

see the attached very familiar demonstration of "special angles"

$(\cos\theta,\sin\theta)$

**Prove It** · January 23rd 2010, 09:05 PM

OK, now that you have found the "focus angle", we now go to a unit circle approach.

A unit circle is a circle of radius 1.

Have a look at the picture below:

For any angle made with the positive $x$ axis (call it $\theta$ ), the vertical distance is $\sin{\theta}$ and the horizontal distance is $\cos{\theta}$ .

If your angle is $60^\circ = \frac{\pi}{3}^C$ , can you see another angle which has the same vertical distance?

**davidman** · January 23rd 2010, 09:17 PM

Ok, I see how there would be two solutions. Thinking in degrees, $180-\theta=\theta_2$

and I googled converting radians/degrees and I must admit that was pretty straightforward...

however, about the second half of the original question,

$sin2\theta=\frac{\sqrt{3}}{2}$

considering the first part of the question, $2\theta=60\:\:, \theta=30$ makes sense the way I see it. But how are there four solutions and not two?

**bigwave** · January 23rd 2010, 09:23 PM

radians x $\left(\frac{180^o}{\pi}\right) =$ degrees

degrees x $\left(\frac{\pi}{180^o}\right)=$ radians

**Prove It** · January 23rd 2010, 09:40 PM

Originally Posted by davidman

Ok, I see how there would be two solutions. Thinking in degrees, $180-\theta=\theta_2$

and I googled converting radians/degrees and I must admit that was pretty straightforward...

however, about the second half of the original question,

$sin2\theta=\frac{\sqrt{3}}{2}$

considering the first part of the question, $2\theta=60\:\:, \theta=30$ makes sense the way I see it. But how are there four solutions and not two?

OK, for any given domain, the number of solutions depends on the period of the function.

Can you see that if you go around the circle (i.e. if you travel $360^\circ$ ) you get to the same angle?

If you travel another $360^\circ$ , you get back to the same angle again.

And so on, and so on...

So if you wanted to show all solutions to

$\sin{x} = \frac{\sqrt{3}}{2}$

then $x = \left\{ 60^\circ, 120^\circ \right \} + 360^\circ n$ , where $n$ is some integer representing the number of times you have gone around the unit circle.

Now if $x = 2\theta$ then

$2\theta = \left \{ 60^\circ, 120^\circ \right\} + 360^\circ n$ .

Now, to solve for $\theta$ we have to divide both sides of the equation by $2$ . This includes the $360^\circ$ ...

So $\theta = \left \{ 30^\circ, 60^\circ \right\} + 180^\circ n$ .

So letting $n = 0$ we have

$\theta = \left \{ 30^\circ, 60^\circ \right \}$ .

If we let $n = 1$ we have

$\theta = \left \{ 210^\circ, 240^\circ \right \}$ .

If we let $n = 2$ we have

$\theta = \left \{ 390^\circ, 420^\circ \right \}$ . These solutions are out of the original $360^\circ$ .

So the four solutions are

$\theta = \left \{ 30^\circ, 60^\circ, 210^\circ, 240^\circ \right \}$ .

**davidman** · January 23rd 2010, 09:46 PM

Originally Posted by Prove It

Awesome explanation.

Ooooooooohhhhhhhhhh, I get it now!

$\theta_1=\frac{\pi}{6}\:,\:\theta_2=\frac{\pi}{3}\ :,\:\theta_3=\frac{7}{6}\pi\:,\:\theta_4=\frac{12} {9}\pi$

Thread: New to trigonometry

LinkBack

Thread Tools

Display

New to trigonometry

Q1,q2

Similar Math Help Forum Discussions

Trigonometry to Memorize, and Trigonometry to Derive

How To Do Trigonometry For This...?

How To Do This Trigonometry

trigonometry

Trigonometry

Search Tags