Double-Angle Formulas...

[Cod]

For some reason, I cannot comprehend the concepts behind this. I read the example problems over and over; however, I still cannot understand the process when I go to study or do work on it.

Just to refresh your minds, the double-angle formulas:

sin2x = 2(sinx)(cosx)
cos2x = cos^2x - sin^2x = 1 - 2sin^2x = 2cos^2x-1
tan2x = 2tanx/1-tan^2x

The book example:

If cosx = -2/3 and x is in quadrant II; find sin2x and cos2x.



If someone could explain the processes when using these formulas to solve problems, I'd greatly appreciate it. The book just isn't helping me any.

[Guybrush Threepwood]

sin x = sqrt(1 - (cos x)^2). Since cos x = -2/3 we have:
sin x = sqrt(5/9). So sin x = sqrt(5)/3 or sin x = -sqrt(5)/3.
We know that x is in the second quadrant and that makes sin x > 0. So sin x = sqrt(5)/3. Now you know sin x and cos x. Just replace them and find sin 2x and cos 2x.

cos 2x = (cos x)^2 - (sin x)^2 [:D] so it's less confusing.....

[dextercioby]

It's quite simple really:you have cos x,then you compute sinx and sustitute in the formulas for the double angle.Got it??

[Cod]

So how would you go about finding 'tan2x'? I understand that tanx = sinx/cosx. I just don't see how you can plug that into the formula: tan2x = 2tanx/1-tan^2x.


Unless...

2(sinx/cosx)/1-(sin^2x/cos^2x) <-----would that be correct?

If that's correct, would I just plug in the known values of sin and cos? Then do the arithmatic?

[Guybrush Threepwood]

yes, you could do that or you could do (sin 2x/cos 2x) after you have found the previous two results.

[AndersHermansson]

I think this might be what you are looking for:

sin(2x) = sin(x+x)
cos(2x) = cos(x+x)
tan(2x) = sin(2x)/cos(2x)

now we use the rule of addition:
sin(x+x) = sin(x)cos(x) + cos(x)sin(x) = 2sin(x)cos(x)
cos(x+x) = cos(x)cos(x) - sin(x)sin(x) = cos^2(x) - sin^2(x) = cos^2(x) - (1 - cos^2(x)) = 2cos^2(x) - 1

tan(2x) = 2sin(x)cos(x) / 2cos^2(x) - 1 ... etc