[Math] Trigonometric ratios for angles greater than $90^\circ$

geometrytrigonometry

The trigonometric ratios of an angle greater than $90^\circ$ are equal to the supplementary angle's ratios.

I'm just clarifying this, but the ratios don't actually exist for angles greater than $90^\circ$ right? (since by definition these ratios are of a right angle triangle). Is it just mathematical custom to assume the ratios of an angle greater than $90^\circ$ to be equivalent to its supplementary angle's ratios? For convenience?

Best Answer

Instead of ratios in right triangles (which as you notice make sense only for acute angles), one can consider the cosine and sine defined as the $x$ and $y$ coordinate of a point that moves around a unit circle. This works for all angles -- and for acute angles you can inscribe a right triangle in the first quadrant of the unit circle and see that the unit-circle definition matches the right-triangle one.

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After 90°, the cosine becomes negative, because the point is now to the left of the $y$ axis (so the $x$ coordinate is negative).

After 180°, the sine becomes negative too -- both coordinates of the moving point are now negative.

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