Simple trigonometry question

From my basic geometry, I remember being able to express certain terms like: sin30 = 0.5 and cos45 = (sqrt2)/2.

Is there a similar, simple way of expressing the sine, cosine and tangent of 15, and the sine and cosine of 75? (I know tan75 = sqrt3 + 2).

Just playing around on a calculator with numbers doesn’t seem to be a particularly efficient way of finding these things out - is there another way that someone can recommend?

You could use some trig identities (Google should bring up a page or two) to look for things like:

sin (t/2) = sqrt((1-cos(t))/2)

so using cos(30) = sqrt(3)/2:

sin(15) = sqrt((1-sqrt(3)/2)/2) = sqrt (1/2 - sqrt(3)/4)

For 15 you could use the half angle formulas

``````   sinx/2=sqrt((1-cosx)/2) and cosx/2=sqrt((1+cosx)/2)
``````

since the cos of 30 deg is (sqrt3)/2 you get

sin 15 deg= sqrt((1-(sqrt3)/2)/2) and similarly for cosine.

For 75deg you could just add the results for 15 deg to the well known values for 60.

However, I think it would be easier to use the inverse trig buttons on your calculator.

I should have known that cursed Race Bannon would beat me to the post! I’ll get you next time, Bannon! and that foolish Quest child too!

Thanks - that’s appreciated. I’d forgotten about identities (it’s been a while…).

Larry, please don’t tell Dr. Quest I’ve been using his computer. I haven’t finished washing the hovercraft yet.

Gotta get those eels out of there, huh?

In general, you can use the addition and subtraction formulas to figure out the functions of an arbitrary angle, but the “weirder” it is, the more unpleasant the process is.

I can’t believe I said “for 75 deg just add the results for 15 with the well known values for 60.” Obviously I should have said use the addition formula for 15 and 60 degs.

Hold up a second there. If you’re saying what I’m thinking you’re saying, then I think you’re wrong. But you may not be saying what I’m thinking you’re saying. So what are you saying?

For instance, you can’t get a real, functional-form expression for sin(10°). I’ve tried. The best you can do is use the addition formula sin(30°) = sin(10° + 10° + 10°). This allows you to determine that sin(10°) is a root of the polynomial 8x[sup]3[/sup] - 6x + 1. This root can be expressed with familiar symbolics, but only as the sum of two complex numbers whose imaginary parts cancel, IIRC.

Would “the sum of two complex numbers whose imaginary parts cancel” not be a “functional-form expression”? Or, what do you mean by a “functional-form expression”?

I agree with your main point, however. Even if we figure out the sine of 10 degrees, there’s still the sine of, say, 12 degrees.

sin(12°) = [sqrt(2sqrt(5) + 10) + sqrt(3) - sqrt(15)] / 8, but I understand what you mean.

Notice I didn’t just say “functional-form expression”. I said “real, functional-form expression”. The OP asked for a “simple way”. This way is, by its nature, complex. I apologize. I shouldn’t have said “wrong”, but I do think it’s misleading.

Hmm…you may be right. It’s been a while since I did stuff like this.

[sub]disabled smilies - DrMatrix[/sub]

Man… did I really do that much winking?

No, you didn’t.

ultrafilter, did you add them?

I’m kidding. It’s a glitch. I disabled smileys in that post so it wouldn’t happen.