Question

Will give medal and fan!

Answer 1

Given tan theta=-sqrt2/2 and theta lies in quadrant IV, find cos theta/2

Answer 2

@SyedMohammed98 @TheSmartOne

Answer 3

@mady111

Answer 4

yes!

Answer 5

You can help me?

Answer 6

i can

Answer 7

Sweet. What do I need to do first?

Answer 8

hold on

Answer 9

\[\sin \theta \sqrt{2/2}\]

Answer 10

\[\cos \theta=-\sqrt{2/2}\]

Answer 11

- in the first on to

Answer 12

\[\tan \theta=1 \cot=1 \csc -\sqrt{2} \sec =-\sqrt{2}\]

Answer 13

since the angle formed in quadrant III is 90 degrees, then if we bisect that angle we'll get half of that which is 45 degrees....
so we know that in a right triangle with 45 degrees, the sides will have a ratio of 1:1:square root of 2
so let's suppose our hypotenuse is sqrt of 2 and the legs have measure of 1. Since we're in quadrant 3 both legs will have measure of -, (because at quadrant 3, x and y are both negative) 1 but the hypotenuse will still be sqrt2... so we'll have

using this info, you can now compute for the 6 trig. functions of theta..
so sin theta=-1/sqrt2=-sqrt(2)/2
cos theta=-1/sqrt2=-sqrt(2)/2
tan theta=sin theta/cos theta <a trigonometric identity
since sin theta=cos theta in this case,
tan theta=1
cot theta=1/tan theta=1/1=1
sec theta= 1/cos theta=1/(-1/sqrt2)=-sqrt2
csc theta=1/sin theta=1/(-1/sqrt2)=-sqrt2

Answer 14

So I use cos theta=-1/sqrt2=-sqrt(2)/2

Answer 15

yes you got it

Answer 16

Do I have to plug in anything for tan?

Answer 17

Or I just put that equation in the calculator?

Answer 18

no put that in the calculator

Answer 19

did you get the answer you was looking for

Answer 20

I put it into wolfram because I don't have a graphing calculator and the answer was exactly what I put in

Answer 21

hmm ill do it a different way for you

Answer 22

Alright

Answer 23

The six trig functions are listed below @ stands for theta. x,y, and r are in reference to the trig functions related to a unit circle.

Cos@ = adj/hyp or x/r....Sec@ = hyp/adj or r/x
Sin@ = opp/hyp.or y/r..Csc@ = hyp/opp or r/y
Tan@ = opp/adj or y/x...Cot@ = adj/opp or x/y
Since your angle is QIV, Xs are positive, Ys are negative
This means that your Sine, Cosecant, Tangent, and Cotangent are going to be negative. With that said...

The sec@ = Sqrt(3)/1 remember any number can be written as a ratio by putting it over a 1. This is the key. x = 1 and r = sqrt(3).

Using the Pythagorean theorem since we form a right triangle in drawing the coordinate system and @ in the 4th Quad, we get...that x = 1, r = sqrt(3) and y is all we need to find.
(1)^2 + y^2 = (sqrt3)^2
y = sqrt(2)

Now you have adj or x = 1, opp or y = sqrt(2), and
hyp or r = sqrt(3)
Write your functions and remember which ones are negative.
Sin@ = y/r = sqrt2/sqrt3 = sqrt6/3 (negative)
Cos@ = x/r = 1/sqrt3 = sqrt3/3 (positive)
Tan@ = sqrt2/1 = sqrt 2 (negative)
Csc@ = r/y = sqrt3/sqrt2 = sqrt6/2 (negative)
Cot@ = x/y = 1/sqrt2 = sqrt2/2 (negative

Answer 24

is this any better?

Answer 25

I'm sorry. I'm still confused. I'm not sure what to do next.

Answer 26

okay ill draw it out for you

Answer 27

Okay thank you