Question

i will give u a medal and a fan!!!!!!!!!
if tan(theta)=-1,and the sin(theta)>0,what is the value of cos(theta)?

Answer 1

@ganeshie8 ????????

Answer 2

Use the identity \(\large {tan(\theta) = \frac{sin(\theta)}{cos(\theta)}}\)

Answer 3

ok,then what?

Answer 4

@mathmate ?

Answer 5

Oh, I thought they gave you the value of sin(theta).
Sorry, now use this:
\[sec^2(\theta)=1+tan^2(\theta)\]
but since \(sec(\theta)=\frac{1}{cos(\theta)}\)
you will be able to solve for cos.

Answer 6

and then what do i do after solving for cos?

Answer 7

That seems to be the answer the question is asking for... unless I'm mistaken again! :)

Answer 8

let me give u the solution of the question:

Answer 9

Since tan(theta) < 0 and sin(theta) > 0, the terminating side of the angle is in the second quadrant.

tan (theta)= -y/x = 1/-1 = -11;
r = square root of (1 ^2 + 1 ^2) = square root of 2;
cos (theta) = negative one over the square root of 2 = negative (square root of 2 over 2)

Answer 10

i know this is really confusing!

Answer 11

Not at all!

Answer 12

how did v get to this solution?can u plz help me?

Answer 13

First, you've followed the right steps.
tan(theta)=-y/x=-1

Answer 14

Since tan(theta)=sin(theta)/cos(theta), AND sin(theta)>0

Answer 15

cos(theta) must be <0.
ok so far?

Answer 16

yes:)

Answer 17

Also, since tan(theta)=-1, so
cos(theta)=-sin(theta), ok there?

Answer 18

yes,u solved for cos theta,right?

Answer 19

Yes, all you need is after solving for cos(theta), put a negative sign on it,
since sin(theta)>0, and cos(theta)=-sin(theta), so cos(theta)<0.

Answer 20

yep,but then how did they find the radius,that is,square root of 2?

Answer 21

As you did earlier, you had
sec^2(theta)=2
so \(cos(\theta)=\pm \sqrt{2}/2\)
and we know that it is negative.

Answer 22

Use the identity
\[sec^2(\theta)=1+tan^2(\theta)=1+(-1)^2=2\]
so
\[cos^2(\theta)=\frac{1}{2}\]

Answer 23

oh,ok.so to solve for cos,we will find the square root of 1/2,right

Answer 24

Yep.

Answer 25

so that's basically it?

Answer 26

Note also that after rationalizing the denominator:
\[\sqrt{\frac{1}{2}}=\frac{\sqrt2}{2}\]

Answer 27

yes,i get it now.Thankyou sooooooooooo much @mathmate !

Answer 28

Yes, that's all to it. When problems are broken down to pieces, they always seem easy! :)

Answer 29

that's right:)

Answer 30

You're welcome! @mayaal !