# The Sine and the Cosine of Angles Greater Than 90^o

• Feb 4th 2011, 08:03 PM
Barthayn
The Sine and the Cosine of Angles Greater Than 90^o
I need help understanding the following questions:

6. The point (20, 21) is on the terminal arm of an angle θ in standard position. Find sin θ and cos θ.

7. The point (-8, 6) is on the terminal arm of an angle θ in standard position. Find sin θ and cos θ.

I thought it was draw the point on the graph and then find the angle from the origin (sin 0). But the teacher said no. This is not what it means. She did not know what it meant. As well, what does it mean by sin θ and cos θ? I have no idea what it means.
• Feb 4th 2011, 08:07 PM
Prove It
You should know that $\displaystyle x= r\cos{\theta}$ and $\displaystyle y = r\sin{\theta}$, where $\displaystyle r$ is the radius (in your case, length of the arm).

So first evaluate $\displaystyle r$ using Pythagoras and your two points $\displaystyle (0,0)$ and $\displaystyle (20,21)$, then you can evaluate $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$.

Do the same for Q7.
• Feb 4th 2011, 08:11 PM
Barthayn
Quote:

Originally Posted by Prove It
You should know that $\displaystyle x= r\cos{\theta}$ and $\displaystyle y = r\sin{\theta}$, where $\displaystyle r$ is the radius (in your case, length of the arm).

So first evaluate $\displaystyle r$ using Pythagoras and your two points $\displaystyle (0,0)$ and $\displaystyle (20,21)$, then you can evaluate $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$.

Do the same for Q7.

So this is asking for the sides of the triangle? Not, the angles? How would we know the sides if we do not have the angle it is rotated by?
• Feb 4th 2011, 08:16 PM
Prove It
I suggest you research the unit circle.

But yes, you have been given enough information to create a right-angle triangle, and you don't need to know anything about the angles.

Draw a set of Cartesian axes and join the origin to the point (20, 21).

Surely you can see that the angle made with the positive x axis (which you don't need to measure) gives the opposite side = 21 units and the adjacent side = 20 units.

Use Pythagoras to find the length of the Hypotenuse and you can find $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$.
• Feb 4th 2011, 08:23 PM
Barthayn
Quote:

Originally Posted by Prove It
I suggest you research the unit circle.

But yes, you have been given enough information to create a right-angle triangle, and you don't need to know anything about the angles.

Draw a set of Cartesian axes and join the origin to the point (20, 21).

Surely you can see that the angle made with the positive x axis (which you don't need to measure) gives the opposite side = 21 units and the adjacent side = 20 units.

Use Pythagoras to find the length of the Hypotenuse and you can find $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$.

I understand what the points mean, they are the lengths of the sides, however, I do not understand what it is meant by sin θ and cos θ. Wouldn't they be the same angle?

P.S. My last post I was a little confused.
• Feb 4th 2011, 08:26 PM
Prove It
Surely you know that $\displaystyle \sin{\theta} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}$ and $\displaystyle \cos{\theta} = \frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}$...

Once you know the length of the Hypotenuse you can find $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$...
• Feb 4th 2011, 08:38 PM
Barthayn
Quote:

Originally Posted by Prove It
Surely you know that $\displaystyle \sin{\theta} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}$ and $\displaystyle \cos{\theta} = \frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}$...

Once you know the length of the Hypotenuse you can find $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$...

So this entire question is not asking you for the angle of the triangle, but for the decimals that it may be.
• Feb 4th 2011, 10:58 PM
Prove It
Quote:

Originally Posted by Barthayn
So this entire question is not asking you for the angle of the triangle, but for the decimals that it may be.

No, it's not asking for the angles.

I don't think you understand exactly what $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$ represent. They represent the ratio of the length of the sides of a triangle. When the angles are the same, these ratios are identical, which is why $\displaystyle \theta$ appears after the sine or cosine.
• Feb 5th 2011, 07:29 AM
Barthayn
Quote:

Originally Posted by Prove It
No, it's not asking for the angles.

I don't think you understand exactly what $\displaystyle \sin{\theta}$ and $\displaystyle \cos{\theta}$ represent. They represent the ratio of the length of the sides of a triangle. When the angles are the same, these ratios are identical, which is why $\displaystyle \theta$ appears after the sine or cosine.

I know they are ratios. What I mean by the decimals is that sin θ = 21/29 and cos θ = 20/29. Are these answers correct?
• Feb 5th 2011, 07:43 AM
Prove It
Yes they are correct. Keep them as fractions.
• Feb 5th 2011, 09:06 AM
Here's a visual guide
• Feb 5th 2011, 10:50 AM
Barthayn
Thanks for the image, But why are there sin(pi-θ) and cos (pi-θ)??? Also why is the question labeled as Angles Greater Than 90^o?
• Feb 5th 2011, 11:20 AM
Quote:

Originally Posted by Barthayn
Thanks for the image, But why are there sin(pi-θ) and cos (pi-θ)??? Also why is the question labeled as Angles Greater Than 90^o?

I wanted to label the angles differently as I placed both on a single diagram.

The $\pi-\phi=180^o-\phi$ is the second angle formed by (0, 0) and (-8, 6)

If we subtract $\phi$ from $180^o$ then we get your second $\theta$ which starts from the positive side of the x-axis anticlockwise,
as does the first $\theta$

From the sketch, you can work with the acute angle shown in the green triangle.

One of the angles that you want is greater than 90 degrees,
so the title refers to that one.

When calculating Sine and Cosine of angles greater than 90 degrees,
we use the horizontal and vertical co-ordinates,

horizontal for Cosine, which is why the numerator has $-8$
for your larger angle,

vertical for Sine, which is why the numerator has $+6$ for the second angle.

When the angles are less than 90 degrees, working with the side lengths
of a right-angled triangle suffices.