# Math Help - [SOLVED] Value of sin2x and cos2x when tanx=5/3

1. ## [SOLVED] Value of sin2x and cos2x when tanx=5/3

if tan x=5/3

a) What is the value of sin (2x)
b) What is the value of cos (2x)

my awnser have to be in fraction

Im not allowed to use a calculater to get an awnser and im unsure what im suppose to do...

I think that sin x=5
and cos x=3
but i really don't know...

Janne the Swede

2. $tanx=\frac{sinx}{cosx}$

hence, $\frac{sinx}{cosx}=\frac{5}{3}$

also, $sin^2x+cos^2x=1$

$(\frac{5cosx}{3})^2+cos^2x=1$

$34cos^2x=9$

$17cos2x+17=9$

hence,
$cos2x=\frac{-8}{17}$

3. Originally Posted by JannetheSwede
if tan x=5/3

a) What is the value of sin (2x)
b) What is the value of cos (2x)

my awnser have to be in fraction

Im not allowed to use a calculater to get an awnser and im unsure what im suppose to do...

I think that sin x=5
and cos x=3
but i really don't know...

Janne the Swede
Hi

Be careful : sin x and cos x are always between -1 and 1

$\sin 2x = 2 \sin x \cos x = 2 \tan x \cos^2 x$

A very useful relationship is $\cos^2 x = \frac{1}{1+\tan^2 x}$

$\sin 2x = 2 \frac{\tan x}{1+\tan^2 x} = \frac{15}{17}$

4. Are you allowed to use these (not so well known) multiple angle formulas?

$\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$

$\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$

If so, just plug in 5/3 for $\tan \theta$ and simplify.

01

5. Originally Posted by malaygoel
$tanx=\frac{sinx}{cosx}$

hence, $\frac{sinx}{cosx}=\frac{5}{3}$

also, $sin^2x+cos^2x=1$

$(\frac{5cosx}{3})^2+cos^2x=1$

$34cos^2x=9$

$17cos2x+17=9$

hence,
$cos2x=\frac{-8}{17}$
$(\frac{5cosx}{3})^2+cos^2x=1$

$34cos^2x=9$

what do you do in between these steps?? i don't get it

6. Originally Posted by JannetheSwede
$(\frac{5cosx}{3})^2+cos^2x=1$

$34cos^2x=9$

what do you do in between these steps?? i don't get it
Multiply both sides of the equation by 9...then recollect like terms

7. Originally Posted by malaygoel
Multiply both sides of the equation by 9...then recollect like terms
Ahhh, i see ^^ but why 9?

Never mind i saw it now

8. I was unable to correctly use your way of getting cos2x

I tested it on another question.

tan(x)=1/7
find a. a*cos(2x)
=>
((1*cosx)/7)²+cos²x=1
2cos²x=49
and i just saw what I did wrong... i forgot to multiply the ...+cos²x with 49.. nm xD

after i got 50cos²x=49
is the awnser then cos2x+1=7
=cos2x=-6/25

or did I do it wrong somewhere?

9. Originally Posted by JannetheSwede
I was unable to correctly use your way of getting cos2x

I tested it on another question.

tan(x)=1/7
find a. a*cos(2x)
=>
((1*cosx)/7)²+cos²x=1
2cos²x=49
and i just saw what I did wrong... i forgot to multiply the ...+cos²x with 49.. nm xD

after i got 50cos²x=49
is the awnser then cos2x+1=7
=cos2x=-6/25

or did I do it wrong somewhere?
$50cos^2x=49$

$50\frac{cos2x+1}{2}=49$

10. so cos2x=((49-1)*2)/50?

=48/25

11. Originally Posted by JannetheSwede
so cos2x=((49-1)*2)/50?

=48/25

Spoiler:

$
50\frac{cos2x+1}{2}=49
$

$
25cos2x+25=49
$

$cos2x=\frac{24}{25}$

12. ok, thanks a lot