Solving for theta in a Trigonometric Equation

[CrossFit415]

Homework Statement

Tan $$\Theta$$ = 12 / 5, sin$$\Theta$$<0

Homework Equations

Sin $$^{2}$$ $$\Theta$$+ Cos $$^{2}$$$$\Theta$$ = 1

The Attempt at a Solution

Sin is less than 0 so... it should be somwhere in the II or III quadrant?

What identities should I use to solve?

 

[HallsofIvy]

If $sin(\theta)$ is negative then $\theta$ is in either the third or fourth quadrants, not the second or third. But since $tan(\theta)$ is positive, that means $cos(\theta)$ is also negative and so $\theta$ is in the third quadrant.

If you were to construct a right triangle with legs of length 12 and 5, what length would the hypotenuse be? If you are trying to find $\theta$ I don't believe you will find any simple value.

 

[CrossFit415]

If $sin(\theta)$ is negative then $\theta$ is in either the third or fourth quadrants, not the second or third. But since $tan(\theta)$ is positive, that means $cos(\theta)$ is also negative and so $\theta$ is in the third quadrant.

If you were to construct a right triangle with legs of length 12 and 5, what length would the hypotenuse be? If you are trying to find $\theta$ I don't believe you will find any simple value.

Ok. So sin < 0 then cos theta turns to a negative also?

 

[CrossFit415]

Tan $$^{2}$$ $$\theta$$ + 1 = sec $$^{2}$$ $$\theta$$

I could use this identity to solve it

 

[icystrike]

Yes u can... But its meaningless

U take reciprocal and use the identity $$sin^{2}(x)+cos{2}(x)=1$$

But u will lose the information whereby tan(x)>0 ...

I would suggest u get the basic angle and just "shift" it to the third quadrant...

 

[QuarkCharmer]

To me, it looks like the most important identity to use in solving this for angle theta is

$$tan\theta = \frac{sin\theta}{cos\theta}$$

Since you know that sine of theta is negative, cos of theta must also be negative. What quadrant would that place the angle theta in? (Considering both the X and Y coordinates are negative). Drop your reference triangle here.