Thread: How Good Am I?
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Gib Z
#61
Jan16-07, 06:47 AM
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Quote Quote by murshid_islam View Post

you used the substitution [tex]x = \frac{1}{2}\tan\theta[/tex]
from this, you get, [tex]\tan\theta = 2x[/tex] and from this, you get [tex]\theta = \arctan(2x)[/tex] and [tex]\sin\theta = \frac{2x}{\sqrt{4x^2+1}}[/tex] and [tex]\cos\theta = \frac{1}{\sqrt{4x^2+1}}[/tex]. and you also know that [tex]\sin 2\theta = 2\sin\theta\cos\theta[/tex]
Sorry, I'm afraid you may be a bit confused. I used the substitution x=1/2 tan theta on Hootenanys approach, When I get to where you were showing me, I was doing my own trigonometric substitution that was not x=1/2 tan theta. SOrry