tan(x) = sin(x)/cos(x) question

[joint52]

tan(x) = sin(x)/cos(x)

Problem: If tan(x) = 5/12, find sin(x) and cos(x)

My answer:

I know this can be solved by drawing a right-angled triangle and using
Pythagoras's theorem to find the hypotenuse. Using this method gives
sin(x) = 5/13, cos(x) = 12/13.

But by looking at tan(x) = 5/12, I'm thinking sin(x) = 5 and cos(x) =
12.

But then sin(x) and cos(x) must be between -1 and 1 so I'm thinking sin
(x) could be 1/4 and cos(x) could be 3/5, because 1/4 divided by 3/5
gives 5/12.

Why am I wrong?

TIA

[Paul Sperry]

In article
<be7af643-8411-4f55...@r5g2000yqb.googlegroups.com>,
joint52 <joi...@yahoo.com> wrote:

There's an x such that sin(x) = 1/4 and there's a y such that
cos(y) = 3/5. Unfortunately, x =/= y.

--
Paul Sperry
Columbia, SC (USA)

[Barry Schwarz]

On Sat, 21 Nov 2009 13:29:57 -0800 (PST), joint52 <joi...@yahoo.com>
wrote:

You are ignoring the additional constraint that
(sin x)^2 + (cos x)^2 = 1

From the original, you know that
5 * (cos x) = 12 * (sin x)

Two equations, two unknowns, ...

--
Remove del for email

[Virgil]

In article <l2sgg59565mhqv3n0...@4ax.com>,
Barry Schwarz <schw...@dqel.com> wrote:

And, in this case at least, two solutions:

sin(x) = 5/13 and cos(x) = 12\13

and

sin(x) = -5/13 and cos(x) = -12\13

[Stan Brown]

Sat, 21 Nov 2009 13:29:57 -0800 (PST) from joint52 <joint52
@yahoo.com>:

> But then sin(x) and cos(x) must be between -1 and 1 so I'm thinking sin
> (x) could be 1/4 and cos(x) could be 3/5, because 1/4 divided by 3/5
> gives 5/12.

Because [sin x]^2 + [cos x]^2 = 1 for every x, but (1/4)^2 + (3/5)^2
does not equal 1.

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai...