# New to trigonometry

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• Jan 23rd 2010, 07:20 PM
davidman
New to trigonometry
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 $\displaystyle sin\theta=\frac{\sqrt{3}}{2}$

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

second part of the question:

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

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

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

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

thanks for any help.
• Jan 23rd 2010, 07:32 PM
Prove It
Quote:

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

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

second part of the question:

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

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

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

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

thanks for any help.

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

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

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

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

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

Which angle (out of $\displaystyle 30^\circ$ and $\displaystyle 60^\circ$) has the Opposite side with length $\displaystyle \sqrt{3}$ units and the Hypotenuse as $\displaystyle 2$ units?
• Jan 23rd 2010, 07:34 PM
VonNemo19
Quote:

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

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

second part of the question:

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

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

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

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

thanks for any help.

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

Well, $\displaystyle \sqrt3/2$ is given by evaluating $\displaystyle \sin\theta$ 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
• Jan 23rd 2010, 07:35 PM
bigwave
Q1,q2
Quote:

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

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

second part of the question:

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

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

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

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

thanks for any help.

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

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

these are quadrants 1 and 2 respectfully
• Jan 23rd 2010, 08:45 PM
davidman
Quote:

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

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

Thanks!

Quote:

Originally Posted by VonNemo19
I think that you mean to say that $\displaystyle \sin\theta=\frac{\sqrt{3}}{2}$ has two solutions for $\displaystyle 0\leq\theta\leq2\pi$ ? And you wish to find the solutions?

Exactly that, yes.

Quote:

Originally Posted by bigwave
$\displaystyle sin\theta=\frac{\sqrt{3}}{2}$

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

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... ?
• Jan 23rd 2010, 09:00 PM
bigwave
you may be looking at a triangle in just one quadrant

see the attached very familiar demonstration of "special angles"

$\displaystyle (\cos\theta,\sin\theta)$
• Jan 23rd 2010, 09:05 PM
Prove It
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:

http://upload.wikimedia.org/wikipedi...angles.svg.png

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

If your angle is $\displaystyle 60^\circ = \frac{\pi}{3}^C$, can you see another angle which has the same vertical distance?
• Jan 23rd 2010, 09:17 PM
davidman
Ok, I see how there would be two solutions. Thinking in degrees, $\displaystyle 180-\theta=\theta_2$

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

however, about the second half of the original question,

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

considering the first part of the question, $\displaystyle 2\theta=60\:\:, \theta=30$ makes sense the way I see it. But how are there four solutions and not two?
• Jan 23rd 2010, 09:23 PM
bigwave
radians x $\displaystyle \left(\frac{180^o}{\pi}\right) =$ degrees

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

http://upload.wikimedia.org/wikipedi...ersion.svg.png
http://en.wikipedia.org/wiki/File:De...Conversion.svg
• Jan 23rd 2010, 09:40 PM
Prove It
Quote:

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

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

however, about the second half of the original question,

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

considering the first part of the question, $\displaystyle 2\theta=60\:\:, \theta=30$ 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 $\displaystyle 360^\circ$) you get to the same angle?

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

And so on, and so on...

So if you wanted to show all solutions to

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

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

Now if $\displaystyle x = 2\theta$ then

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

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

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

So letting $\displaystyle n = 0$ we have

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

If we let $\displaystyle n = 1$ we have

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

If we let $\displaystyle n = 2$ we have

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

So the four solutions are

$\displaystyle \theta = \left \{ 30^\circ, 60^\circ, 210^\circ, 240^\circ \right \}$.
• Jan 23rd 2010, 09:46 PM
davidman
Quote:

Originally Posted by Prove It
Awesome explanation.

Ooooooooohhhhhhhhhh, I get it now! (Nod)

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