# Solving Problems With Trigonometry

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Yes, the sine, on the first period, takes on the value of I've done the algebra; that is, I've done the factoring and then I've solved each of the two factor-related equations. So now I can do the trig; namely, solving those two resulting trigonometric equations, using what I've memorized about the cosine wave.

From the first equation, I get: However (and this is important!

Now I'll use the reference angles I've memorized to get my final answer.

Note: The instructions gave me the interval in terms of degrees, which means that I'm supposed to give my answer in degrees.

So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees.

So all we need to do is-- well we can simplify the left-hand side right over here. So angle W plus 155 degrees is equal to 180 degrees.

So I'll give you a few seconds to think about how to figure out what a is. So if we're looking at angle Y, relative to angle Y, this is the opposite. Well if we don't remember, we can go back to Soh Cah Toa. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So we can divide both sides by that, by cosine of 65 degrees.

These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So we now know that this has length 10.7, approximately. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Well we see from Soh Cah Toa cosine deals with adjacent over hypotenuse.

), I squared to get this solution, and squaring is an "irreversible" process. If you square something, you can't just square-root to get back to what you'd started with, because the squaring may have changed a sign somewhere.) So, to be sure of my results, I need to check my answers in the , which is the same almost-solution as before.

After doing the necessary check (because of the squaring) and discarding the extraneous solutions, my final answer would have been the same as previously.

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