Question

how do you use the unit circle to find the values of the six trig functions for each angle 45 degrees

Answer 1

Read:
http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png


Then remember:
SOH : \(\sin \theta = \huge \frac{opposite}{hypotenuse} \)

CAH : \(\cos \theta = \huge \frac{adjacent}{hypotenuse} \)

TOA : \(\tan \theta = \huge \frac{opposite}{adjacent} \)


And finally:
Cosecant = csc( ) = inverse of sine
Secant = sec( ) = inverse of cosine
Cotangent = cot( ) = inverse of tangent

Inverse means 1 divided by that stuff.
Example:
\(X\) inverse \(\rightarrow \huge \frac{1}{X}\)


Best answer? ;D

Answer 2

okay i gave you best answer.. now just do the problem for me cause im confused! six trig functions for the angles 45

Answer 3

..please

Answer 4

Do you see the 45 degree angles listed in the image?

Answer 5

yes sirr

Answer 6

Cosine is your x component of the coordinate there
Sine is your y component coordinate there

Secant is that cos(45\(^o\)) number flipped, inverse.
Cosecant is that sin(45\(^o\)) number flipped, inverse again.

Tangent is sine divided by cosine
\[tan \theta = \frac{cos \theta}{sin \theta}\]

Can you divide this?
\[\huge \frac{(\frac{\sqrt{2}}{2})}{(\frac{\sqrt{2}}{2})} = \frac{\sqrt{2}}{2}*\frac{2}{\sqrt{2}}\]

And cotangent is the inverse of tangent, 1/tan

Answer 7

We're supposed to help others learn, not just straight up answer the problem. The moderators frown on that: http://openstudy.com/code-of-conduct

:-)

So I'm hoping this is making sense

Answer 8

yeah i want to learn, im just getting frustrated.

Answer 9

Ack I had a typo, it's supposed to read:
\[tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{rise}{run}\]

Please ignore the previous part in the above post.

Answer 10

I can make errors too :-)

Answer 11

i think that yall are just so good in this subject, that when you try to explain to others its very overwelming

Answer 12

i know this might be alot to ask but if you copuld take me step by step if you can, if you cant because you dont have enough time i understand

Answer 13

I can think the same thing about Calculus 3 ;-) But you have to keep breaking down the step until it makes sense to you, something that requires a bit of self-discipline to do. And self-discipline for something that can get crazy complicated if you're not following the steps like in Calculus (and most Calculus professors will skip dozens of steps at a time to save how long it would take to write it all out) gets a whole lot harder. I've struggled with that myself. It's part good instruction, and part self-motivation to try it out, from my experience.

I am a bit limited on time today. The key thing here to notice is patterns. Cos(angle) typically gives you the x-coordinate (for how we typically orient things in class), sin(angle) typically gives you the y-coordinate. That wikimedia image I linked for you right at the start of this question is your key.

All you have to do to read cosine or sine of any angle off that is to remember the coordinates are basically ( cos(angle) , sin(angle) ).

So for example, for an angle of 0 degrees you have (1, 0), cosine is 1, sine is 0. Makes sense because you're 100% horizontal, and 0% vertical.

For tangent, it's sine divided by cosine, rise of run, the slope if you will. Do the arthritic for the fraction I posted above with the square roots in it, see if things cancel (they do).

The inverse functions cosecant, secant, and cotangent are just those flipped, inverted, 1 divided by that.

Answer 14

\[\sin \theta = \sin(45^o) = \frac{opp}{hyp} = \frac{\sqrt{2}}{2} \approx 0.07071...\]

Answer 15

Thaanks A lot and i was going to ask does that square root of 2 over 2 cancel out the other one? iif that makes since

Answer 16

\[\cos \theta = \cos(45^o) = \frac{adj}{hyp} = \frac{\sqrt{2}}{2} \approx 0.07071...\]

Answer 17

yay! honestly you doing the problem just made me remember how to do it!!!

Answer 18

^_^