Finding Exact Values for tan^2θ=2tanθsinθ on the Interval 0 ≤ θ ≤ 2π

[rought]

Homework Statement

Solve for 0 ≤ θ ≤ 2π (pi). Give the exact values.

tan^2θ=2tanθsinθ


I'm not sure how to go about this one =/

 

[Dunkle]

First off, observe that this equation holds when tan(\theta) = 0, so you should be able to find three solutions immediately.

Now, try and find more solutions. Hint: you can safely divide by tan(\theta) now because you know it's not equal to 0. Then, remember that
tan(\theta) = \frac{sin(\theta)}{cos(\theta)}.

Let's see what you can do from there!

 

[rought]

ok I worked it out and i got: cosθ=0 and cosθ=1/2 which gave me π/2 and π/3 ... does that seem right?

 

[Dunkle]

The cos(\theta) = 1/2 is correct, but I don't know how you arrived at cos(\theta) = 0. If cos(\theta) = 0, then tan(\theta) is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for cos(\theta) = 1/2 on [0,2\pi], one of them is \pi/3 as you mentioned. What do you think the other one is? Drawing a graph out might help.

 

[rought]

The cos(\theta) = 1/2 is correct, but I don't know how you arrived at cos(\theta) = 0. If cos(\theta) = 0, then tan(\theta) is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for cos(\theta) = 1/2 on [0,2\pi], one of them is \pi/3 as you mentioned. What do you think the other one is? Drawing a graph out might help.

the other one would be 5π/3

here's what I did

tan^2θ=2tanθsinθ
sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
1/cos^2 θ=2/cosθ
2cos^2 θ-cosθ=0
cosθ(2cosθ -1)=0

cosθ=0 and 2cosθ -1=0

 

[Dunkle]

Ok, I see what you did now. The reason that doesn't work is because, as I mentioned in my previous post, tan(\theta) is undefined when cos(\theta) = 0. So, this is not a valid answer.

Here's what I was attempting to get you to do:

tan^{2}(\theta) = 2tan(\theta)sin(\theta)<br /> =&gt; tan(\theta) = 2sin(\theta)<br /> =&gt; \frac{sin(\theta)}{cos(\theta)} = 2sin(\theta)<br /> =&gt; \frac{1}{cos(\theta)} = 2<br /> =&gt; cos(\theta) = \frac{1}{2}<br />

You can divide by tan(\theta) and sin(\theta) because we assume that tan(\theta) \ne 0 because we already found the solutions when tan(\theta) = 0.

Is this clear?

 

[qntty]

I'm not sure how to go about this one =/

Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.

the other one would be 5π/3

here's what I did

tan^2θ=2tanθsinθ
sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
1/cos^2 θ=2/cosθ
2cos^2 θ-cosθ=0
cosθ(2cosθ -1)=0

cosθ=0 and 2cosθ -1=0

\cos{\theta} doesn't work because tangent is undefines when cosθ=0. You also divided by sinθ, which assumes that \sin{\theta}\not= 0. This caused you to lose that solution which works.

 

[rought]

ah ok thanks a ton guys =]