Question

I'm really stuck, Please Help Me!
If sin = -2/3, which of the following are possible?
(There could be more than one answer)

Answer 1

\[\sqrt(\sin^2 \theta+\cos^2 \theta) = 1\]

Answer 2

\[\tan \theta = \frac{ \sin \theta }{ \cos }\]

Answer 3

So i just plug it in right ?

Answer 4

Yes, we know that sin is -2/3, so just plug in the cosine values.

Answer 5

sin^-1(-2/3)=-41.81

Answer 6

Actually you could just do
\[\sin^2\theta+\cos^2\theta = 1\]

Answer 7

cos(-41.81)=square root 5 /3

Answer 8

and
\[\sec \theta = \frac{ 1 }{ \cos \theta }\]

Answer 9

so it can be A and D ?

Answer 10

positive sign

Answer 11

so just D?

Answer 12

Yeah, not D.

Answer 13

The VERY FIRST thing you should notice is that \(\sin(\theta) < 0\) puts \(\theta\) in Quadrant III or Quadrant IV.

Answer 14

Oh its not D, okay Im getting a little confused !

Answer 15

it is DDDDDDDDDDDDDDDDDDDDDDD

Answer 16

Check B as well.

Answer 17

So B and D ? because it can be B

Answer 18

@DoYourHomework and @yahya90

Answer 19

\[\sin \theta = -2/3; \cos \theta = \sqrt(5)/3; \tan \theta = -2/(\sqrt(5))\]
Not D.

Answer 20

A and B.

Answer 21

Okay thank you @DoYourHomework

Answer 22

D & B

Answer 23

Sorry it was A and B but thank you guys for all your help though I really really appreciate it ! (:

Answer 24

\[\sin \theta = \frac{ -2 }{ 3 }\]
\[\cos \theta = \frac{ \sqrt(5) }{ 3 }\]
\[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\]
\[\tan \theta = \frac{ -2 }{ 3 }* \frac{ 3 }{ \sqrt(5) } = \frac{ -2 }{ \sqrt(5) }\]

Answer 25

In Quadrant III, \(\tan(\theta) > 0\;and\;\cos(\theta) < 0\)
In Quadrant IV, \(\tan(\theta) < 0\;and\;\cos(\theta) > 0\)

Therefore, tangent or cotangent vs. cosine or secant MUST have opposite signs.

This eliminates Choice D.

A right triangle with the sine of some angle equal to 2/3, makes the cosine of that same angle \(\dfrac{\sqrt{3^{2} - 2^{2}}}{3} = \dfrac{\sqrt{9 - 4}}{3} = \dfrac{\sqrt{5}}{3}\;or\;-\dfrac{\sqrt{5}}{3}\)

This eliminates Choice C.

Systematic approaches can be quite helpful.