Find the values of the six trigonometric functions of theta: cos θ = 35/37 constraint: tan θ < 0

- anonymous

Find the values of the six trigonometric functions of theta: cos θ = 35/37 constraint: tan θ < 0

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- anonymous

|dw:1401422933563:dw|

- dan815

six trignometric functions?

- anonymous

yes. sin, tan, sec, cos, cot, csc

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## More answers

- dan815

ok plug into calculator using the inverse buton

- anonymous

to find all six? Can you baby it down for me haha

- dan815

okay lol

- dan815

so first look at your triangle its 35 in x and -37 in y

- dan815

so enter 35/-37 in your calculator, and then do cos-1 (35/-37)

- dan815

thats a button on your calculator cos inverse

- dan815

do the same for sin and tan

- anonymous

cos=2.811295299?

- dan815

make sure you are using degree mode

- anonymous

ok it is now cos=161.0753556

- dan815

that answer is in radians which is fine too if your teacher wants that

- dan815

alright thats one, now do the rest

- dan815

remember that
1/sin= csc
1/cos=sec
1/tan=cot

- dan815

by the way that is not the only solution,

- anonymous

ya it usually accepts fractions.

- dan815

|dw:1401423483117:dw|

- dan815

it found that angle, but if your teacher is particular on you to only find tht last angle on the bottom

- anonymous

So i just plugged in -37/\[\sqrt{2594}\]

- anonymous

is it in quadrant 2 not 4?

- dan815

i dont know its your choice what do you want??

- anonymous

well the cos is positive and the tan is negative. so wont that be quad 4?

- dan815

yeah

- anonymous

ok now what would the hypotenus equal?

- dan815

do Pythagorean thm

- dan815

|dw:1401423730828:dw|

- dan815

ok

- anonymous

i get square root of 2594

- anonymous

is that right? I might be doing my calcs wrong

- dan815

looks about right

- myininaya

|dw:1401423880441:dw|

- anonymous

so would sin of theta=37/square root of 2594

- dan815

the hypotenuse should be roughly sqrt2*36

- anonymous

it says it is wrong

- anonymous

square root of 36?

- dan815

its -37

- dan815

no nvm that, its just an approximation i dont have calc

- myininaya

cos is adj/hyp
you labeled the initial triangle a bit wrong

- dan815

okay do it over

- anonymous

can you show me the right triangle?

- dan815

|dw:1401424044893:dw|

- myininaya

good

- myininaya

now find the opp measurement of the angle theta |dw:1401424168943:dw|

- anonymous

and how did you get that?

- myininaya

use the theorem we talked about last question

- myininaya

you are given cos is 35/37

- anonymous

right right right. got it

- anonymous

none of those are negative?

- myininaya

right cos is positive not negative

- myininaya

sin is negative

- myininaya

so you will definitely get a negative number for that opp measurement to theta

- anonymous

i got 12

- myininaya

and it should really be -12

- anonymous

35^2+b^2=37^2

- anonymous

b^2=144 b=12?

- myininaya

or b=-12
we will go with b=-12 because we were given cos is + tan is - which makes sin -

- anonymous

ok. so sin =-12/37

- myininaya

|dw:1401424409906:dw|
that symbolizes the triangle is to the right of zero but below 0 (that is talking about the signs)

- myininaya

and yes

- anonymous

that clarifies completely. what about this one?
tan θ is undefined. constraint: π ≤ θ ≤ 2π

- anonymous

find the six.

- dan815

remember tan theta = sin theta/costheta
so its undefin when cos theta= 0 which is at?

- anonymous

so each one would be undefined?

- dan815

no just the tan and the cot

- dan815

well atually no cot is fine

- anonymous

how do i find the rest?

- dan815

have to first find this

- dan815

tan theta= sin theta/cos theta => undefined when denominator =0
where cos theta =0

- dan815

solve for cos theta = 0 , when theta is in that given interval pi to 2pi

- anonymous

would it be 2pi/pi?

- dan815

no

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