If tan(theta) = 3/4 and sin(theta) < 0, find cos(theta)?

Question

If tan(theta) = 3/4 and sin(theta) < 0, find cos(theta)

Don't know much trig. Points to answer with best explanation. Thanks

Answer 1

tan(theta) = sin(theta) / cos(theta)

... right? That means, because tan(theta) is positive and sin(theta) is negative, cos(theta) must be negative. Now we put this into a triangle.

Imagine a right-angled triangle with shorter sides of length 3 and 4. What will the hypotenuse be? Exactly: 5. Think about where the angle theta is inside this triangle, and think about what its sin and cos values will be. You should come up with the solution that sin(theta) = 3/5 and cos(theta) = 4/5. Unfortunately, those aren&#039;t quite right, so we have to just multiply them by -1, to get the answer:

cos(theta) = -4/5.

What we really wanted was a triangle with negative side lengths. Even though those sides were negative, you still add their squares to find the hypotenuse, and the hypotenuse would still have been 5. tan(theta) would also still have been 4/5. It&#039;s just that sin and cos would have been negative like they should have been.

Answer 2

tan = y / x, and it is positive in either quadrant 1 or quadrant 3

if sin &lt; 0, that means you&#039;re in quadrant 3, and both x and y are negative (so y = -3 and x = -4)

now use x^2 + y^2 = r^2

x = -4 and y = -3, so r = 5

cos (theta) = x / r = -4 / 5

I think the easiest way to solve these problems is to use the basic definitions of the trig functions, combined with the fact that x^2 + y^2 = r^2

cos = x / r

sin = y / r

tan = y / x

Answer 3

Quadrant 3

cos Ó¨ = 4 / 5