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I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $\theta>90^\circ$? I get that when $\theta>90^\circ$, you find the acute angle that describes $\theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $\cos\theta=\sin(\frac{\pi}{2}-\theta)$ still hold true here?

N. F. Taussig
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Ethan Chan
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  • Use the difference identity and see why. – Sean Roberson Jul 19 '18 at 07:29
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    The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold... – Angina Seng Jul 19 '18 at 07:30
  • @LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold? – Ethan Chan Jul 19 '18 at 07:31
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    You should be formatting your mathematical expressions with MathJax. – N. F. Taussig Jul 19 '18 at 07:41
  • Do you know the definition of trigonometric functions for a general angle $\theta$ which may be positive or negative and less than or greater than $\pi/2$? Based on such a definition you can establish the addition formulas and get all such identities. – Paramanand Singh Jul 19 '18 at 07:58
  • @EthanChan The definition takes a unit circle. Any point on the circumference of the circle has coordinates $(a,b)\quad 0\leq a,b\leq1$. Take a point $(x, y)$ on its circumference which makes an angle $\theta$ with the origin and positive x-Axis. Then $\sin\theta=y,\cos\theta=x$. – DynamoBlaze Jul 19 '18 at 09:06
  • @EthanChan: You may find this answer of mine helpful. – Blue Jul 19 '18 at 11:52
  • Please tell us your definition of sine and cosine. What is their domain of definition? We can't answer your question until we know this first. – Somos Jul 19 '18 at 13:54

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Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,\,0)$, and rotate that radius through an angle $\theta$ anticlockwise so its end becomes $(\cos\theta,\,\sin\theta)$. You could have got the same result by rotating the radius at $(0,\,1)$ clockwise through $\frac{\pi}{2}-\theta$, so $(\cos\theta,\,\sin\theta)=(\sin(\frac{\pi}{2}-\theta),\,\cos(\frac{\pi}{2}-\theta))$.

J.G.
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Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$\cos x=\sin(\frac\pi{2}-x)$$ for $0<x<\frac\pi2$.

Then one would learn the Maclaurin series of $\sin$ and $\cos$, and then one can show that they have an infinite radius of convergence.

Since $(0,\frac{\pi}2)\subset\mathbb C$ and $\cos x=\sin(\frac\pi2-x)$ in this interval, by identity theorem this immediately implies the two functions($\cos x$ and $\sin(\frac\pi2-x)$) are equal for any $x\in \mathbb C$.

Szeto
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