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sin(theta)=-4/5
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How did you get that triangle?

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I drew a triangle and use the fact the sin (theta)=-4/5
remember sine is opposite over hypothenuse

Ohhh, thats how you did it, okay!

Now the 5 trig funtions of theta? I thought there were 3 trig functions o.O

now let's find adjacent side value using pythagorean theorem
a^2+4^2=5^2
a^2=25-16
a^2=9
a=3

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now we can see what cos , and tan are
cos is adjacent over hypotenuse
cos(theta)= 3/5

Tangent is sin(theta)/cos(theta)
-4/5
------ =-4/3
3/5

Okay, i got those two, what are the other 3 they are looking for?

csc(theta)= 1/sin(theta)= 1 = -5/4
-----
-4/5

sec(theta)= 1/cos(theta)= 1/ (3/5) = 5/3

cot(theta)= 1/tan(theta)= 1/(-4/3) =-3/4

Oh we use the recriprocoals

yep

tan(theta)=2=2/1; Quadrant 1
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use pythagorean theorem to find hypotenuse

Okay, so basically, if its a whole number, we put it as a fraction?

yes

Hypotenuse is square root of 3

sqrt[2^2+1^2]=sqrt[5]

Srry, did it in my head, got my trusy calculator :D

Okay, i got it, thanks for helpin :D

anything for my shrink

LOL, i might just pass algebra 2

I took algebra 2 in sophomore year, and it was big deal way back

Im taking it freshman :p

smartie

I wish, lol, i just barely pass every time.

pre calc next year and then calculus

arghhh..dont remind me :/ Im taking summer school to prep for it.

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