Find sin, cos, and tan given sin 2x

[axrw]

"Find sin x, cos x, tan x under the given conditions:

sin 2x = 1/5

pi/2 <= 2x < pi"

I don't really even get where to start.

Thanks for any help.

 

[axrw]

Ok, I've been thinking about this a bit longer and came up with something.

I'm given sin 2x, but I want sin x (and the other two). I first found the sides to a triangle that has 1 as its hypotenuse, and 1/5 and [2*sqrt(6)]/5 as its legs.

Then I can use the half angle identities: sin 2x/2.

I came up with :

sin x = sqrt ([1/2 - sqrt(6)/5]/2)

Unfortunately the answer isn't in the back with this one, but am I on the right track? I don't know what it is, but I have no faith in myself with trig.

Thanks.

 

[tkhunny]

You're making it a little harder than it needs to be. You can use a right triangle with an angle identified as 2x, opposite leg 1, hypotenuse 5 and adjacent leg 2sqrt(6). It substantially simplifies the arithmetic.

You can use the half-angle identities. I tend to remember the double andgle identities more readily.

1) sin(2x) = 2sin(x)cos(x)

2) cos(2x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)

And this one.

3) sin^2(x) + cos^2(x) = 1

That's enough ammunition.

#3 will get you cos(2x)

#2 will get you sin(x) and/or cos(x) from cos(2x)

#3 will get you cos(x) and/or sin(x)

 

[axrw]

Thanks for the reply tkhunny. That is a bit easier.