trig, tan, sin, cos, etc help...10 points easy if u r one of those ppl who understand this!!?

Question

http://i106.photobucket.com/albums/m273/starrynight_03/trigchartcopy.jpg

help me to fill in the blanks of the table, i seemed to have gotten everything right except for the ones in blue, can u tell me what the sec theta, csc theta, etc equals when its at that certain radian and degree?

please help, i need this chart completed in order to do my homework assignment because i need to use these identities.

10 POINTS TO WHOEVER HELPS ME FILL IN THE BLANKS!!

o yea...and my teacher said i had it wrong because i had a square root in my denominator o.O

so can u fill in the blank so that the denominator does not have a square root

Answer 1

tan 30= √3/3
sec 30= 2√3/3
cos 60=1/2
cot 60= √3/3
tan 150=-√3/3
sec 150=-2√3/3
tan 210=√3/3
sec 210=-2√3/3
csc 240=-2√3/3
cot240=√3/3
csc 300=-2√3/3
cot 300=-√3/3
tan 330=-√3/2
sec 330=2√3/3

Answer 2

tan 30= √3/3
sec 30= 2√3/3
cos 60=1/2
cot 60= √3/3
tan 150=-√3/3
sec 150=-2√3/3
tan 210=√3/3
sec 210=-2√3/3
csc 240=-2√3/3
cot240=√3/3
csc 300=-2√3/3
cot 300=-√3/3
tan 330=-√3/2
sec 330=2√3/3

I copied this out of my trig chart. Don't forget that, if you know tan, sin and cos already, cot, cos and sec are simply reverses. And, if you already know the basic identities (0-90), it is just a matter of adding to figure out the rest.

Answer 3

I know it's not filling in the blanks, but for future reference, you can do these questions really easily.

Sec(ant) is the inverse function of cosine, so where you have cos{pi/4} = sqrt2/2, sec{pi/4} is just 2/sqrt2, which is sqrt2.
Csc is the inverse of sine, and the same applies, and cot is the inverse of tan, so where you have cot{pi/6} = sqrt3, tan{pi/6} = 1/sqrt3

Also, to remove the square root sign from the denominator, multiply the top and bottom by the square root (which is the same as multiplying by one)
Eg. 2/sqrt2 = 2*sqrt2 / sqrt2*sqrt2 = 2sqrt2/2.

Answer 4

tan pie/6 = [(3)^(1/2)]/3 ..... (3 divided by root 3)
sec pie/6 = 2[(3)^(1/2)]/3 .... (2 root 3 divided by 3)
cosec pie/3 = sec pie/6
cot pie/3 = tan pie/6
csc 2pie/3 = sec pie/6
cot 2pie/3 = - ( tan pie/6)

sorry... too long and tedious to type


steven's got ur answer for u. now to remove the root form the denominator, just multiply both numerator and denominator by the same root.

Answer 5

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Answer 6

(square root 3/3),(2 square root 3/3),(2square root3/3),(square root 3/3),(2 square root 3/3),(square root 3/3), (square root3/3),(2 square root 3/3),(- square root3/3),(-2 square root 3/3),(-2 square root 3/3),(-square root3/3),(- 2 square root 3/3),(-square root 3/3),(- square root 3),(2) there are all of your answers in order. I hope I can help. You're doing good though. Trig isn't all that fun anyways but keep up the good work!

Answer 7

regarding the root in the denominator, you just have to rationalize.

example: if your answer is 2/sqroot of 5 you have to multiply it by sqroot of 5 / sqroot of 5 to take out the root at the denominator, but still have the same value when simplified. so your answer will be 2squareroot of 5/ 5

looks messed up, but yea thats basically it :))

Answer 8

first of all, sec(theta) = 1/cos(theta)
and csc(theta) = 1/sin(theta)
and cot(theta) = 1/tan(theta)

tan (pi/6) = 1/√3 = (√3)/3
sec(pi/6) = 2/√3 = (2√3)/3

csc(pi/3) = 2/√3 = (2√3)/3
cot(pi/3) = 1/√3 = (√3)/3

csc(2pi/3) = 2/√3 = (2√3)/3
cot(2pi/3) = -1/√3 = (-√3)/3

tan(5pi/6) = -1/√3 = (√3)/3
sec(5pi/6) = -2/√3 = (-2√3)/3

csc(4pi/3) = 2/√3 = (2√3)/3
cot(4pi/3) = 1/√3= (√3)/3

csc(5pi/3) = -2/√3 = (-2√3)/3
cot(5pi/3) = -1/√3 = (-√3)/3

tan(11pi/6) = -1/√3 = (-√3)/3
sec(11pi/6) = 2/√3 = (2√3)/3

that's all the values

Answer 9

having a root in the denominator is acceptable. It just isnt standard mathematical syntax/grammar... whatever you want to call it. But there isnt anything mathematically wrong about it. Its like leaving a fraction unreduced, or calling a negative number in the denominator wrong - its arbitrary.

Answer 10

I've uploaded a complete one for you here:

http://upload.wikimedia.org/wikipedia/en/f/fe/Trig_Table.pdf