Question

If sin Θ = -√3/2 and 3π/2 < Θ < , what are the values of cos Θ and tan Θ?

Answer 1

This requires the use of a unit circle. If you do not already have one, just google it real quick. Let's start by solving for theta. To do this, we need to use arcsin. The arcsin finds at what radian is a number the sin. In our case, we want to know at what radian sinx = -sqrt3/2.
\[\sin^{-1} \frac{ -\sqrt{3} }{ 2} = \theta\]
So at what radian is the sine value -sqrt3/2? I would say sin7pi/6 = -sqrt3/2, but 7pi/6 is not greater than 3pi/2. So, theta must be 5pi/3.
Now we can solve the rest of the equation. costheta = the x-value of theta on the unit circle. So,
\[\cos \frac{5 \pi}{3} = \frac{ 1 }{ 2 }\]
Knowing that
\[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\]
We can solve for tan 5pi/3:
\[\tan \frac{5 \pi}{3} = \frac{ \frac{ -\sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2} } = \frac{ -\sqrt{3} }{ 2 } \times \frac{ 2 }{ 1 } = -\sqrt{3} \]
Hope this helps! If you have any questions, just let me know!

Answer 2

That is so helpful, but none of my answers involve negate square root of 3...

Answer 3

Are you saying it's a multiple choice question? What are your answers then?

Answer 4

cos Θ =-1/2 ; tan Θ =square root of 3

cos Θ =-1/2 ; tan Θ = -1

cos Θ = square root of 3/4; tan Θ = -2

cos Θ =1/2 ; tan Θ = square root of 3

Answer 5

If theta is supposed to be less than 3pi/2 instead of greater than, the first answer would be it. Your formatting is a bit hard to understand in your initial question. If theta is supposed to be greater than 3pi/2, just pick D for now and talk to your teacher about it. If I'm correct, they will fix the issue. If I'm wrong, then they can help you with the question. It's a win-win! Hope everything works out :)