I have very little experience writing proofs so I don't know how to begin.
I recognize that the statement is always true, but I can't go about proving it without using circular reasoning.
How could I write a proof for this?
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2It depends where you start and level of rigour required. If you have this identity, it's straight forward to show formally: $\cos(a + b) = \cos a \cos b - \sin a \sin b$. But I think an intuitive approach going back to the unit circle is more instructive. – Simon S Aug 22 '15 at 23:19
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1If you have geometry: why is the cosine of one of the acute angles in a right triangle equal to the sine of the other? – Ian Aug 22 '15 at 23:19
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Given the sin, or cos, you can take this as the definition of the other :P – Paolo Leonetti Aug 22 '15 at 23:24
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(This is for a calculus 1 class, in a trig review section) – Bassinator Aug 22 '15 at 23:29
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See this answer for intuition. – pjs36 Aug 22 '15 at 23:34
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As noted, you need definitions for $\sin$ and $\cos$. Without that, you can prove nothing at all about them. – GEdgar Aug 23 '15 at 01:46
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This answer may be helpful. – Blue Aug 23 '15 at 02:13
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We can prove this using the well known identity $$\cos (a-b) = \cos a\cos b + \sin a\sin b$$ Here $a = \dfrac{\pi}{2}, b = x$ so now we have \begin{align*} \cos\left(\frac{\pi}{2}-x\right) &= \cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x \\ &=0\cdot\cos x + 1\cdot \sin x \\ &= \sin x \end{align*}
Rick
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Geometrically:
given a right triangle with angles $A$, $B$, and $C=\pi/2$ and sides $a$, $b$, and $c$ opposite the respective angles, you have $$\sin(A)=\frac{a}{c}$$ and $$\cos(B)=\frac{a}{c}$$ And we know that $$B=\pi-A-\pi/2$$ so the result is immediate.
Using Euler's formula:
We know that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, so that $$e^{i(\pi/2-x)}=\cos(\pi/2-x)+i\sin(\pi/2-x)$$ But, this is the same as $$e^{i(\pi/2-x)}=e^{i\pi/2}e^{-ix}=i(\cos(-x)+i\sin(-x))\\ =i\cos(x)+\sin(x)$$
Since two complex numbers are the same only when their real and imaginary parts are respectively equal, the result holds.
Terra Hyde
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