#3**+10 **

Great work Chris and Geno.

I'm going to hit it a different way.

$$\\\tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\\\\

\tan(2x)=\frac{2tanx}{1-tan^2x}\\\\\\$$

so draw a right angled triangle and make one of the acute angles = 2x

the opposite side is $$2tanx=2*2=4$$

the adjacent side is $$1-tan^2x=1-2^2=1-4=-3$$ (the negative sign just means it will not be in the first quad )

Using pythagoras the hypotenuse will be

$$\\h=\sqrt{4^2+(-3)^2}\\

h=\sqrt{16+9}\\

h=\sqrt{25}\\

h=5\\$$

so

$$\\cos(2x)

=\frac{adj}{hyp}

=\frac{-3}{5}

=\;-0.6$$

Melody
Oct 7, 2014

#1**+5 **

If tan(x) = 2, the x = 63.435°.

cos(2x) = cos(2·63.435°) = cos(126.870°) = -0.6

This can also be done without a calculator, if you need that explanation, please repost.

geno3141
Oct 6, 2014

#2**+10 **

If tan x = 2, then y / x = 2 / 1

Then, using, √(x^{2} + y^{2} ) = r^{ }

√(2^{2} + 1^{2}) = √5 = r

So sin x = y/r = 2/√5

And using

cos 2x = 1 - 2(sin x)^{2}

cos 2x = 1 - 2(2/√5)^{2}

cos 2x = 1 - 2 (4/5)

cos 2x = 1 - 8/5

cos 2x = -3/5 = -(.6) .....as geno found !!!!!!!

CPhill
Oct 6, 2014

#3**+10 **

Best Answer

Great work Chris and Geno.

I'm going to hit it a different way.

$$\\\tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\\\\

\tan(2x)=\frac{2tanx}{1-tan^2x}\\\\\\$$

so draw a right angled triangle and make one of the acute angles = 2x

the opposite side is $$2tanx=2*2=4$$

the adjacent side is $$1-tan^2x=1-2^2=1-4=-3$$ (the negative sign just means it will not be in the first quad )

Using pythagoras the hypotenuse will be

$$\\h=\sqrt{4^2+(-3)^2}\\

h=\sqrt{16+9}\\

h=\sqrt{25}\\

h=5\\$$

so

$$\\cos(2x)

=\frac{adj}{hyp}

=\frac{-3}{5}

=\;-0.6$$

Melody
Oct 7, 2014