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Prelim 3U Math Trig Question - Help (1 Viewer)

Thread starter may22
Start date Jul 16, 2021

may22

Well-Known Member

#1

My entire class and teacher tried this proving question about 6 times, on both LHS and RHS, but we were stumped. Any ideas?

D

Drongoski

Well-Known Member

#2

$\frac {sin2\theta - cos2\theta + 1}{sin2\theta+cos2\theta - 1} = \frac {2sin\theta cos\theta + 2sin^2\theta}{2sin\theta cos\theta - 2sin^2\theta}= \frac {2sin \theta(cos\theta + sin \theta)}{2sin\theta (cos\theta - sin\theta)} =\frac {\frac{cos\theta}{cos \theta} + \frac {sin\theta}{cos \theta}}{\frac {cos\theta}{cos\theta} - \frac {sin \theta}{cos\theta}} \\ \\ = \frac {1 + tan\theta}{1-tan\theta} = \frac {tan\frac{\pi}{4} + tan\theta}{1-tan\frac{\pi}{4}tan\theta} = tan (\theta + \frac {\pi}{4})$

Last edited: Jul 16, 2021

may22

Well-Known Member

#3

@Life'sHard @Drongoski Thanks guys!

I'm just not sure what you did in the second last step? How did you go from 1-tan(theta) in the denominator to 1 - tan pi/4 tan(theta)?

Last edited: Jul 16, 2021

may22

Well-Known Member

#4

may22 said:
@Life'sHard @Drongoski Thanks guys!

I'm just not sure what you did in the second last step? How did you go from 1-tan(theta) in the denominator to 1 - tan pi/4 tan(theta)?

Oh wait! It's because tan pi/4 = 1, so, literally... 1 x tan(theta)...lol got it now

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