Proof of Angles going beyond 90 degrees in Trigonometry

Asked Apr 27 '20 at 14:21

Active Apr 28 '20 at 22:57

Viewed 264 times

Since we’re introduced to trigonometric ratios in terms of opposite, perpendicular and hypotenuse, all of which are a part of a right angled triangle. This defines the ratios for angles greater than 0 and less than 90.

For angles greater than 90, we drop a perpendicular from the end-point of the moving arm of the angle and we take the acute angle made by the end-point, the pivot point and the point on the x axis where the perpendicular is dropped.Like Done here

My question being; The way we calculate angles greater than 90, is it defined or derived? If it is defined, then what is the intuition behind defining it like that and if it is derived , proof (and how it’s connected to 0 to 90 part).

I do realise that these ratios were never really meant to be specific to the triangle.But then again, they were introduced that way. Also can you please recommend some books which explain trigonometry in great detail?

Thank you!

edited Apr 28 '20 at 21:03

asked Apr 27 '20 at 14:21

siddharth kashyap

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See this answer of mine for some thoughts on the subject. – Blue Apr 27 '20 at 14:27
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It is an extension of the original definition. There are many ways to perform this extension. – Allawonder Apr 27 '20 at 15:11
Perhaps you should start with how "Angles going beyond 90 degrees" themselves are defined. For example, you might think of these as rotations in two dimensions, or in some other terms. Once a framework for these angles themselves is accepted, Readers will be better able to connect them to values of the usual trigonometric functions. – hardmath Apr 27 '20 at 17:19
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Does this answer your question? Proof of Angles going beyond 90 deg in Trigonometry – amd Apr 29 '20 at 01:05
yup it sure does – siddharth kashyap Apr 29 '20 at 11:46

Proof of Angles going beyond 90 degrees in Trigonometry

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