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# Trig question

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I was wondering how you get fractions out of trig functions? I have a problem which requires me to compute the sin, cos, and tan of PI/4 (Find the x,y on a unit circle that corresponds to the given number). Easy enough PI/4 = .7854 sin(.7854) = .0132 But when I go to check my answer I see sin(PI/4) = sqrt(2)/2. Ok.. But sqrt(2)/2 does not = .0132, it evaluates to .7071... Yes, this is for school. Im taking summer classes (pre-calculus) trying to catch up. But being summer, my professor has limited office hours which conflict with my schedule, and the books (poor) explanation doesn''t account for this.. And it''s not like it''s for a grade, he doesn''t even collect assignments. In fact if anyone could point me at a good overview of trig functions with the unit circle, dealing with radians it''d be quite grateful, because obviously im not understanding alot.. this just happens to be the oddity that presents itself at this moment.

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Standard C functions take radians, not degrees. Multiply your value by (PI/180) before passing it.

*is surprised he got by that without a STFW or RTFM...*

Later,
ZE.

//email me.//zealouselixir software.//msdn.//n00biez.//

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BTW, sin(PI/4 radians) is equal to 0.707... sin(PI/4 degrees) is equal to 0.0137... Normally, values in terms of PI are in radians.

Later,
ZE.

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quote:
Original post by ZealousElixir
Standard C functions take radians, not degrees. Multiply your value by (PI/180) before passing it.

His angle is already in radians. He should either multiply it by 180/PI to obtain degrees, or set his calculator to ''radians'' mode.

Cédric

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quote:
Original post by ZealousElixir
*is surprised he got by that without a STFW or RTFM...*

If only my question had anything to do with c functions.

But yeah, I feel stupid about the deg vs radians thing, thanks. This is the first time I've done anything with radians beyond just converting between the two, don't have the option to just use degrees anymore.

Ok, so how do they get sqrt(2)/2 as the answer to sin(PI/4) instead of just .7071?

I know this seems stupid, but i didn't take math seriously in high school and i'm now playing catchup. Sorry and thanks.

[edited by - absolutdew on July 22, 2002 11:10:38 PM]

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sin(angle) = opposite / hypotenuse in a right angled triangle.

If one angle of the right-angled triangle is pi/4, then you have a right-angled, isosceles triangle - and you can figure out the ratios of the sides with pythagoras'' theorem.

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quote:
Original post by AbsolutDew
Ok, so how do they get sqrt(2)/2 as the answer to sin(PI/4) instead of just .7071?

To extend BerrHunter''s explanation a little further:

There are two fundamental triangles which have interesting trigonometric relationships. These are the right-isosceles triangle and the equilateral triangle.

Consider an isosceles triangle with a right angle between the two sides of equal length. Assume that their length is 1 unit. Then the hypotenuse of this triangle has a length of sqrt(2) units. The two other angles, being equal, must be 45 degrees each (or PI/4 radians). The ratio of side lengths will be the same for any scaled version of this triangle and the angles will remain constant (Theory of Similar Triangles).

Now consider the equilateral triangle where each side is of length 2 units. Each corner angle is 30 degress. Draw the triangle with a flat base and split it in two down the middle, so that the top angle is split into two equal 30 degree angles. Now take one of these smaller triangles. Its base is 1 unit, its hypotenuse is 2 units and its height is sqrt(3) units. Again, the ration of side lengths will be the same for any scaled version of this triangle and the angles will remain constant.

Remembering that sin = opposite/hypontenuse, cos = adjacent/hypotenuse and tan=opposite/adjacent, you can now see why the sin, cos or tan or certain angles is given in terms of integers, or the surds sqrt(2) and sqrt(3).

I hope this helps clarify things for you.

Cheers,

Timkin
So now you can see why certain values of

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thanks i think i understand, thanks to that, and some sleep.

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