Trig

S

shooterman

Junior Member
Joined
Aug 20, 2009
Messages
57
Find the 6 Trig funtions for 135 degree angle

I know that how to find the 6 trig funtions I would need to identify P(x,y) or P(cos t,sin t).
At 135 degrees this would land the point with a reference angle of 35 degrees in quadrant II. The origin is 1 unit.
What I need help on is how my teacher gets ( - sqrt(2)/2, sqrt (2)/2 ) for the P(cos t, sin t).
 
shooterman said:
Find the 6 Trig funtions for 135 degree angle

I know that how to find the 6 trig funtions I would need to identify P(x,y) or P(cos t,sin t).
At 135 degrees this would land the point with a reference angle of 45 degrees in quadrant II. The origin is 1 unit.

What I need help on is how my teacher gets ( - sqrt(2)/2, sqrt (2)/2 ) for the P(cos t, sin t).
Click to expand...

You made a mistake in calculating your reference angle. Can you do it after the correction?

Thanks for showing work.
 
180°-135°=45° are you sure? closest x-axis to this angle is the negative
axis which is at 180°
 


Yes, we're sure that 180 - 135 = 45. (Is that what you're trying to ask?)

You typed 35.

Subhotosh edited your typing, showing the correction in red.

The question, "Are you sure?" is too vague to answer. (Are we sure of what?)

You stated that you're unsure over how your instructor arrived at the values for sine and cosine.

Can you be more specific?

If you understand that the reference angle is 45 degrees (and your use of 35 degrees is a typographical error), then are you asking how we determine that the sine and cosine of a 45 degree angle is sqrt(2)/2?

In other words, are you trying to determine sin(45 degrees) and cos(45 degrees) by hand?

I'm not sure where you're stuck, in your given exercise.

If you're unable to find the words to describe why you're stuck, then can you at least show us what you're trying, instead?

Perhaps, if you show your work, we'll be able to determine the stage at which you've reached the "obstacle".

We can't read your mind. 8-)

 
I guess am asking how to finding the length of the legs and hypotenuse from the degrees by useing cos t, sin t,
 
shooterman said:
… how to [find] the length of the legs and hypotenuse from the degrees by [using] cos t, sin t …
Click to expand...


It looks (to me) like you're thinking backwards. (However, IF you're not typing what you're actually thinking, then that's an entirely different issue.)

We don't use the value of sin(45°) or cos(45°) to determine measurements of a reference triangle.

Instead, we use measurements of a triangle to determine the value of sin(45°) or cos(45°).

ACTUALLY, in exercises like this one, we've already MEMORIZED the values for sin(45°) AND cos(45°), and we're using 45° as a reference angle for 135°.

The sine and cosine of a 45-degree angle are both the same.

sin(45°) = sqrt(2)/2

cos(45°) = sqrt(2)/2

If you're trying to ask how to manually determine these two values using right-triangle trigonometry, then say so, and I will show you how.

However, if you're trying to ask how to use a reference angle to determine the sine and cosine of 135°, then say so, and I will show you how.

I still do not understand exactly what you're trying to do (i.e., where you're at in this particular exercise) because you still have shown no work.

Please, let us know. 8-)

 
"If you're trying to ask how to manually determine these two values using right-triangle trigonometry"
Yes, right-triangle as in 90-45-45 degree 90-60-30. I am sure this is what i am looking for.
 


Okay, I think that I now understand what you're trying to ask.

Read the information at THIS PAGE.

If there's anything you don't understand, then please come back here and post specific questions.

If you find the informationat that page too confusing, then let me know, and I'll try to provide some background information and/or other examples.

Cheers 8-)

 


Great!

Here's an example of how you could have phrased the question in your original post:

"How can I find sin(45 degrees) and cos(45 degrees) using a right triangle?"

:p

 
You must log in or register to reply here.
Top