Thread: Analytic Trigonometry help...

Hi,

Today we started working on what our book calls, Analytic Trigonometry. I am quite lost as to how I solve these problems. My college has a "Math Lab," but they were extremely busy when I went during my break between classes. I wasn't able to get any help there before I had to take off for my next class.

The first is (I will post some more after I figure this out if I still need the help):

2sin^2 theta -1 = 0

So I add one to the RS and divide by two leaving me with...

sin^2 theta = 1/2

Then I take the sqrt of both sides and get:

sin theta = 1/sqrt(2).

That is where I am lost. On some of the problems I am able to look at the unit circle to find the answer I am looking for, but what is 1/sqrt(2) on the unit circle? Some are simple, like the sin of sqrt of 1/2...

Then after I find that number I always seem to have issues getting the points. In this case I would draw the sin curve and put a line at 1/2.

Thanks in advance for the help! It is greatly appreciated!


-Tom

You're looking for angle, theta, with the property that sin(theta)=1/sqrt{2}.

This happens to be pi/4. You just have to be familiar with certain values of the 30-69-90 and 45-45-90 triangles. Lot's of times the angle can't be expressed in a closed form by hand.

Originally Posted by TomCat

Hi,

Today we started working on what our book calls, Analytic Trigonometry. I am quite lost as to how I solve these problems. My college has a "Math Lab," but they were extremely busy when I went during my break between classes. I wasn't able to get any help there before I had to take off for my next class.

The first is (I will post some more after I figure this out if I still need the help):

2sin^2 theta -1 = 0

So I add one to the RS and divide by two leaving me with...

sin^2 theta = 1/2

Then I take the sqrt of both sides and get:

sin theta = 1/sqrt(2).

That is where I am lost. On some of the problems I am able to look at the unit circle to find the answer I am looking for, but what is 1/sqrt(2) on the unit circle? Some are simple, like the sin of sqrt of 1/2...

Then after I find that number I always seem to have issues getting the points. In this case I would draw the sin curve and put a line at 1/2.

Thanks in advance for the help! It is greatly appreciated!


-Tom

If the answer is not obvious to you, use a calculator

sin(theta) = 1/sqrt(2)
theta = arcsin(1/sqrt(2)) = 45 degrees

to get the answer and then draw the picture to gain familiarity.

But are you really supposed to figure these out without a calculator?

No. I believe that we shouldn't be using a calculator.

I'm still stumped though. What I'm trying to do is find where the sin theta =1/sqrt(2) line intercepts the sin curve I believe.

Here's what we did in class...

tan^2 theta = 1/3

tan theta = + or - 1/sqrt(3).

Then we graph tan and draw the lines 1/sqrt(3) and - 1/sqrt(3). Where those two lines intercept the graph, from 0 to 2pi are what I'm looking for.

So theta = pi/6. Then I have to add pi to that and get 7pi/6. That was for the positive...

The negative is -pi/6, so add pi and get 5pi/6 and add pi to that and get 11pi/6.

So the intercepts are {pi/6, 5pi/6, 7pi/6, 11pi/6} .

That one is very simple to me.

Would the intercepts for the problem I posted above be {pi/4, 5pi/5}. My prof did not do a good job explaining this at all...

Thanks again!


-Tom