# Math Help - trigonometric identity question

1. ## trigonometric identity question

the questions is to find all possible measures of angle $\theta$ if $0\leq\theta\leq360^o$
$sin\theta= - 1/\sqrt2$

and

$tan\theta = -1$

I know they are both in the 4th quadrant and in that quadrant sin and tan are both negative, but where do i go from there?
What is the rules and steps to solve other problems similar to this?

2. Originally Posted by crosser43
the questions is to find all possible measures of angle $\theta$ if $0\leq\theta\leq360^o$
$sin\theta= - 1/\sqrt2$

and

$tan\theta = -1$

I know they are both in the 4th quadrant and in that quadrant sin and tan are both negative, but where do i go from there?
What is the rules and steps to solve other problems similar to this?
$\tan \theta = \frac{\sin \theta}{\cos \theta}$

since tan and sin are negative that means cosine is positive, so we must be in the fourth quadrant.

further, you can find exactly what cosine is equal to:

$\cos \theta = - \sin \theta = \frac{\sqrt{2}}{2}$

At this point you want to check with your trig unit circle. One of the first google hits is:
http://dcr.csusb.edu/LearningCenter/...UnitCircle.gif

Good luck!!

3. how would i solve for $tan\theta=-1$

4. In the unit circle, $\cos \theta = x$ and $\sin \theta = y$. As previously mentioned, $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$. Look at the unit circle and find the point where the x- and y- coordinates are the same but with opposite signs.

BTW, the way I read the problem, the two trig ratios are separate problems. If that is the case, then Quadrant IV isn't the only place where sine or tangent is negative. Sine is also negative in Quadrant III and tangent is negative in Quadrant II. Sure, they aren't the same quadrant, but if these ratios are separate problems, that shouldn't matter.

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