Thread: More double angle and Half angle problems

tanx = -5/12; x in quad IV

so I got tan 2x = -120/119 but I don't understand how to get sin2x and cos2x.

cosx = 5/9; x in quad I

Like the previous problem, I am able to get cos2x, which is -31/81, but I don;t know how to get the other 2 functions.

Lastly,
sinx = 2sin x/2 cos x/2
I'm not really sure hot to prove this. It looks like a double angle sin identity.

Originally Posted by hellojellojw

tanx = -5/12; x in quad IV

so I got tan 2x = -120/119 but I don't understand how to get sin2x and cos2x.

cosx = 5/9; x in quad I

Like the previous problem, I am able to get cos2x, which is -31/81, but I don;t know how to get the other 2 functions.

Lastly,
sinx = 2sin x/2 cos x/2
I'm not really sure hot to prove this. It looks like a double angle sin identity.

$\displaystyle \tan{x}=\frac{\sin{x}}{\cos{x}}$...

Originally Posted by VonNemo19

$\displaystyle \tan{x}=\frac{\sin{x}}{\cos{x}}$...

Well, I'm not that dumb. I did try that, but I got the wrong answer.
Unless I'm doing it wrong. Given tanx = -5/12 sinx = -5 cosx = 12
Right????

Originally Posted by hellojellojw

Well, I'm not that dumb. I did try that, but I got the wrong answer.
Unless I'm doing it wrong. Given tanx = -5/12 sinx = -5 cosx = 12
Right????

Not quite... I never said you were dumb.

$\displaystyle -1\leq\sin{x}\leq1$

Same goes for cosine.

Use the pythagorean theorem to find the hypotenuse (h) (where -5 and 12 are the legs, respectively). then $\displaystyle \sin{x}=\frac{-5}{h}$

Originally Posted by VonNemo19

Not quite... I never said you were dumb.

$\displaystyle -1\leq\sin{x}\leq1$

Same goes for cosine.

Use the pythagorean theorem to find the hypotenuse (h). then $\displaystyle \sin{x}=\frac{-5}{h}$

ahahaha I guess I am dumb. so sinx = -5/13. Unless I'm still doing it wrong.

Originally Posted by hellojellojw

so sinx = -5/13.

Indeed!