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.