Trig identity: How do I know that cos^2(x) + sin^2(x) = 1 for ALL angles?

apple2357

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Mar 9, 2018
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The standard way of proving cos squared plus sine squared equals 1 is to use a right angled triangle. That's the approach I see in a textbook. But how do I know it works for all angles?
 

Dr.Peterson

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Nov 12, 2017
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What definition are you using for the trig functions of non-acute angles?

If you use the unit-circle definition (that the sine and cosine of θ are, respectively, the x-coordinate and the y-coordinate of the point on the unit circle \(\displaystyle x^2 + y^2 = 1\) at angle θ from the positive x-axis), then the equation of the circle becomes the identity!
 

apple2357

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Mar 9, 2018
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So if we define sine and cosine in terms of ratio of sides in right angled triangles then we are restricted with the identity. But if we use sine and cosine defined by the x and y coordinate on a unit circle, then we don't have to worry about the issue? So why is it often introduced by the former rather than the latter?
 

Otis

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… So why is [trig] often introduced by [right-triangle definitions] rather than the [unit circle]?
One reason could be that beginning trig students have more experience working with right triangles (eg: finding lengths using Pythagorean formula) than they do finding coordinates of points on a circle. That situation seems similar to measuring angles in degrees first, then introducing radian measure later. That is, we start with simpler concepts and work up to more complicated ones.

😎
 

ksdhart2

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Mar 25, 2016
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This is almost certainly overkill, but if you use the complex definitions for sine and cosine, the identity falls into place.

\(\displaystyle sin(z) = \frac{e^{iz} - e^{-iz}}{2i}\)

\(\displaystyle cos(z) = \frac{e^{iz} - e^{-iz}}{2}\)

\(\displaystyle sin^2(z) = \left( \frac{e^{iz} - e^{-iz}}{2i} \right)^2 = -\frac{e^{-2iz} + e^{2iz} - 2}{4}\)

\(\displaystyle cos^2(z) = \left( \frac{e^{iz} + e^{-iz}}{2} \right)^2 = \frac{e^{-2iz} + e^{2iz} + 2}{4}\)

\(\displaystyle sin^2(z) + cos^2(z) = \frac{(e^{-2iz} + e^{2iz} + 2) - (e^{-2iz} + e^{2iz} - 2)}{4} = \frac{4}{4} = 1\)
 
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