Quick maths question!July 30, 2006 3:42 AM   Subscribe

Quick maths [very basic trigometry] question.

If cos(x) - sin(x) = 1/4, what is x, and therefore cos(x) * sin(x)? How do I go about working this out? I'm highly embarrassed that I can't work this out myself. Thanks!
posted by PuGZ to Education (9 answers total)

Best answer: square both sides of the equation, then use the fact that cos(x)cos(x) + sin(x)sin(x) = 1 to solve for cos(x)sin(x) directly.

I'm not sure (my brain is tired,) but I don't think there is a single solution for x.
posted by blenderfish at 3:52 AM on July 30, 2006

Assuming, then you used the above to establish a (known) K such that sin(x)cos(x) = K,

Use the trig identity:
sin s cos t = (sin(s + t) + sin(s - t)) / 2

so, s and t are both X,

K = sin(x)cos(x) = (sin(2X) + sin(0)) / 2

sin(0) is 0.

K = sin(x)cos(x) = sin(2x) / 2
K = sin(2x)/2
2K = sin(2x)

sin^-1(2K) = x

so, that should be it.
Of course, there are infinitely many solutions, but you can pick the one closest to 0 if you want.
posted by blenderfish at 4:06 AM on July 30, 2006

Last post, I promise..

The one I used above is listed as 'obscure' on the page I linked to.

The more important trig identity to remember (i.e., the one you should probably remember for the duration of the class, even if you forget it the day after the final) is:
sin (s + t) = sin s cos t + cos s sin t

from which, if you take S and T to both be X, you get:
sin 2X = 2 sin X cos X

also, remember
cos (s + t) = cos s cos t – sin s sin t

okay; I'm out.
posted by blenderfish at 4:14 AM on July 30, 2006

cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

Put y = pi/4; then cos(y) = sin(y) = sqrt(2)/2 = 1/sqrt(2)

Substituting:

cos(x+pi/4) = cos(x)/sqrt(2) - sin(x)/sqrt(2)
sqrt(2)cos(x+pi/4) = cos(x) - sin(x)

Given cos(x) - sin(x) = 1/4:

sqrt(2)cos(x+pi/4) = 1/4
cos(x+pi/4) = 1/(4*sqrt(2)) = sqrt(2)/8

x + pi/4 = 2*pi*n +/- arccos(sqrt(2)/8), n integer
x = 2*pi*n +/- arccos(sqrt(2)/8) - pi/4, n integer
posted by flabdablet at 5:04 AM on July 30, 2006

Best answer: cos(x) - sin(x) = 1/4:
(cos(x) - sin(x))2 = 1/16
cos2(x) + sin2(x) - 2sin(x)cos(x) = 1/16

But

sin2(x) + cos2(x) = 1 (identity), so

1 - 2sin(x)cos(x) = 1/16
2sin(x)cos(x) = 15/16
sin(x)cos(x) = 15/32
posted by flabdablet at 5:13 AM on July 30, 2006

Response by poster: Thanks to both of you! Flabdablet's last answer made the most sense to my highly-embarrassed high-school student self, so I marked that as the best answer. This is a fantastic resource!

(I say that far too often!)
posted by PuGZ at 6:10 AM on July 30, 2006

I was terribly pleased with myself for figuring out how to special-case cos(x+y) = cos(x)cos(y) - sin(x)sin(y) to fit your problem, and would be glad to explain the steps further if they're not clear.

Or were you in fact only interested in cos(x)sin(x), and not actually in need of x itself? If so, I think blenderfish deserves some marks as well, since all I did was work his method through for completeness.

Trig is a lot easier when you have Google to remember the identities for you :-)

Blenderfish, I really want to thank you for linking to the page with the quick explanation of prosthaphaeresis. I had heard about a non-logarithm-based method of doing logarithm-like things, but had never actually seen it laid out before.
posted by flabdablet at 7:15 AM on July 30, 2006

Or were you in fact only interested in cos(x)sin(x), and not actually in need of x itself? If so, I think blenderfish deserves some marks as well, since all I did was work his method through for completeness.

Yeah, I didn't want to do the exact math, because I was very tired and worried I would screw it up.

It's interesting that I solved for X by first going to the sin(x)cos(x), while you solved for X directly from the cos(x) - sin(x) form; I used the sin(x+y) identity, and you used the cos(x+y) identity; I substituted X for both variables, while you substituted X for one, and (cleverly) a particular value for y for the other.

The fact you can go totally different routes and still get the same correct answer is one of the neatest things about math.

I really want to thank you for linking to the page with the quick explanation of prosthaphaeresis.

Yep, no problem. Ran into that by accident; I thought that was cool, too. It's hard for people my age to even believe there was once a world without pocket calculators.
posted by blenderfish at 2:31 PM on July 30, 2006

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