Double-Angle Formulas

  • Thread starter Cod
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  • #1
Cod
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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.
 

Answers and Replies

  • #2
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 so it's less confusing.....
 
  • #3
dextercioby
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It's quite simple really:you have cos x,then you compute sinx and sustitute in the formulas for the double angle.Got it??
 
  • #4
Cod
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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?
 
Last edited:
  • #5
yes, you could do that or you could do (sin 2x/cos 2x) after you have found the previous two results.
 
  • #6
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
 
Last edited:

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