+0

# Tan x=2 ,cos2x=?

#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

$$\\cos(2x) =\frac{adj}{hyp} =\frac{-3}{5} =\;-0.6$$

.
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.

Oct 6, 2014
#2
+10

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

Then, using, √(x2 + y2 ) = r

√(22 + 12) = √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   !!!!!!!   Oct 6, 2014
#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
#4
+5

Here's yet another way:

tanx = 2 implies sinx = 2cosx  ...(1)

cos2x = cos2x - sin2x.     Using (1) this is  cos2x = -3cos2x  ...(2)

sin2x + cos2x = 1.     Using (1) this is 5cos2x = 1   which means that  cos2x = 1/5   ...(3)

Put (3) in (2) to get   cos2x = -3/5

Oct 7, 2014
#5
0

Thanks,

It is great when we all weigh in like this.

I like Chris's and Alan's ways best but they are all good. Oct 7, 2014