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Use of Radian Measure (1 Viewer)

Thread starter 205021
Start date Mar 1, 2010

2

205021

New Member

#1

Hey, I'm really struggling with this stuff, so theres probably more to come.

Prove:

tan (θ + π/4) tan (θ - π/4) = -1

P

philly2000

New Member

#2

$LHS = \tan(\theta +\frac{\pi}{4})\tan(\theta-\frac{\pi}{4}) \\ = (\frac{\tan\theta + \tan\frac{\pi}{4}}{1-\tan\theta\tan\frac{\pi}{4}})(\frac{\tan\theta - \tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}) \\ = \frac{\tan^2\theta - 1}{1-\tan^2\theta} \\ = -\frac{-\tan^2\theta +1}{1-\tan^2\theta} \\ = -1 \\ = RHS \\$

2

205021

New Member

#3

philly2000 said:
$LHS = \tan(\theta +\frac{\pi}{4})\tan(\theta-\frac{\pi}{4}) \\ = (\frac{\tan\theta + \tan\frac{\pi}{4}}{1-\tan\theta\tan\frac{\pi}{4}})(\frac{\tan\theta - \tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}) \\ = \frac{\tan^2\theta - 1}{1-\tan^2\theta} \\ = -\frac{-\tan^2\theta +1}{1-\tan^2\theta} \\ = -1 \\ = RHS \\$

thanks for that. just a question, how did you get from line 2 to line 3?

edit: nah i got it. got pi confused, thought it equalled 90 degrees.

Last edited: Mar 1, 2010

2

205021

New Member

#4

And I hold true to my word. this one please:

Simplify: 2cos^(2)3x - 1 where x = 2pi/9

thanks!

xV1P3R

Member

#5

$2cos^2(3x)-1=cos(6x)..............($Cosine double angle$) \\=cos(\frac{6(2\pi)}{9}) \\=cos(\frac{4\pi}{3}) \\=-cos(\frac{\pi}{3})............($cosine negative in the 3rd quadrant$) \\=-0.5$

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