If cos(t)=−9/11 where pi

Question

If cos(t)=−9/11 where pi<t<(3pi/2), find the values of the following trigonometric functions.

cos(2t) =
sin(2t)=
cos(t/2)=
sin(t/2)=

anonymous · Answer

Find first sin(t)

$$\sin(t) = \sqrt{1 - \cos^2(t)}$$

Now find :

$$\sin(2t) = 2 \sin(t) \cdot \cos(t)$$

anonymous · Answer

For cos(2t) use :

$$\cos(2t) = 2\cos^2(t) - 1$$

poofypenguin · Answer

how do you find sin(t) in exact value and no decimals?

anonymous · Answer

We can find it..

Do not simplify the square root;;

Like :

$$\sin(t) = \sqrt{1 - \frac{81}{121}} \implies \sin(t) = \sqrt{\frac{40}{121}} \implies \sin(t) = \frac{2 \sqrt{10}}{11}$$

As interval is given as pi<t<(3pi/2), here sin(t) is negative so:

$$\sin(t) = -\frac{2 \sqrt{10}}{11}$$

poofypenguin · Answer

I understand it so far! Thanks for explaining!
So, how would i go about solving the half ones?

anonymous · Answer

$$\cos(\frac{t}{2}) = 2\cos(t) - 1$$

poofypenguin · Answer

I've never seen that identity before...is there one for sine too?

anonymous · Answer

It is double angle formula only..

anonymous · Answer

I think I forgot the square there sorry...

ajprincess · Answer

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Wont it be this @waterineyes?