Question

Find the exact value of the remaining trigonometric functions of theta.

tan theta = 2*6^1/2, cos theta < 0

Answer 1

\[\tan \theta = 2\sqrt{6}\]

Answer 2

\[\theta =\tan^{-1} \left( 2\sqrt{6} \right)\]
\[\theta =78.463 +n.\Pi\]

Answer 3

why you selected these 2 func. ?

Answer 4

I need to find the EXACT value of sin theta, cos theta, cot theta, sec theta, and csc theta. But if anyone can help me get one of them, I think I can figure out the rest.

Answer 5

Thanks!

Answer 6

you can not h3

Answer 7

one is easy: \[\cot \theta = \left( 1 \over \tan \theta \right) \]

Answer 8

so
\[\cot \theta = \left( 1 \over 2.\sqrt{6} \right)\]

Answer 9

This has become very confused.

Answer 10

and Theta is in the third quadrant

Answer 11

Yes, I got that.

Answer 12

i got something: i use the t formulas for getting sin theta

Answer 13

\[\tan \theta = \left( 2.t \over 1-t ^{2} \right)\]

Answer 14

substitute
\[\tan \theta = 2.\sqrt{6}\]

Answer 15

and solve for t, that is a quadratic equation ( http://nl.wikipedia.org/wiki/Vierkantsvergelijking)

Answer 16

determinant is 100, and t1 and t2 are
\[t1 =\left( 3 \over \sqrt{6} \right)\ t2 = \left( 2 \over \sqrt{6} \right)\]

Answer 17

then i get for cos theta
\[\cos \theta = \left( 1-t ^{2} \over 1+t ^{2}\right) \]

Answer 18

which gives me
\[\cos \theta = \left( -1\over 5 \right) and \cos \theta =\left( 1 \over 5 \right)\]

Answer 19

there is only one possible solution:
cos theta < 0 as in the question so
\[\cos \theta = \left( -1\over 5 \right)\]

Answer 20

Perhaps I can try from the t formulas...thanks for your help.

Answer 21

last solution must be correct:
\[\cos \theta = \left( -1\over 5 \right)\]

Answer 22

similarly i get the sin theta:
\[\sin \theta = \left( 2.t \over 1+t ^{2} \right)\]

Answer 23

with
\[t1 =\left( 3 \over \sqrt{6} \right)\ t2 = \left( 2 \over \sqrt{6} \right)\]

Answer 24

\[\sin \theta =\left( 12 \over 5 \sqrt{6} \right)\]

Answer 25

I get (4*sqrt 6)/25

Answer 26

Oh, I was supposed to divide by 2 to get t, right?

Answer 27

if i use the two solutions of the quadratic equation t1 and t2 i get for sin theta same answer.

Answer 28

\[\sin \theta = \left( 2.t \over 1+t ^{2} \right)= \left( \left( 6 \over \sqrt{5} \right)\over \left( 1+\left( 9 \over 6 \right) \right) \right) = \left( 12 \over 5.\sqrt{6} \right)\]

Answer 29

\[\sin \theta = \left( 2.t \over 1+t ^{2} \right)= \left( \left( 6 \over \sqrt{6} \right)\over \left( 1+\left( 9 \over 6 \right) \right) \right) = \left( 12 \over 5.\sqrt{6} \right)\]

Answer 30

isn't t defined as tan X = X/2?

Answer 31

Rather, t would be X/2 given tan X?

Answer 32

for t1, and
\[\sin \theta = \left( 2.t \over 1+t ^{2} \right)= \left( \left( 4 \over \sqrt{6} \right)\over \left( 1+\left( 4 \over 6 \right) \right) \right) = \left( 24 \over 10.\sqrt{6} \right)\]
for t2, which is the same result

Answer 33

different tangent, that is the point with the t formulas, you dont need to know what t stands for.

Answer 34

now sec and csc are just resp. 1/cos and 1/sin

Answer 35

and sin theta is negative too:
\[\sin \theta = -\left( 12 \over 5.\sqrt{6} \right)\]

Answer 36

Thanks!