How can I algebraically prove this trigonometric symmetry for sin?
$\sin(-x) = -\sin(x)$
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The angle $-\theta$ is the angle $\theta$ reflected through the $x$-axis. (I assume you're using "right-angle trig".) – David Mitra Jan 29 '14 at 14:42
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Yes, but can we prove this algebraically? – George Newton Jan 29 '14 at 14:43
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1It drops out of applying the difference law to $\sin(0-x)$. (I don't recall if the difference law uses this fact in its proof, though.) – David Mitra Jan 29 '14 at 14:45
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Define "algebraically prove". Using $\sin x=\sum\cdots$ is OK? – Martín-Blas Pérez Pinilla Jan 29 '14 at 14:46
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3It depends on how you define the sine in the first place. That identity might well be by definition, for instance, if you define sines geometrically. – egreg Jan 29 '14 at 14:58
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Are you fine with assuming $\cos$ is even? :) – user76568 Jan 29 '14 at 14:59
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What level are you working at? If you know series, think about the series representation of the functions. – Michael Dyrud Jan 29 '14 at 14:43
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You can go from the definition of $sin(x) = \sum^\infty_{n=0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$
From there, factor $-x$ into $(-1)x$ and use $(xy)^a = x^ay^a$
Notice since $n$ is an integer you can say that $(-1)^{2n+1} = (-1)^{1}= -1$
kleineg
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If you can use complex numbers and Euler's formula, try this algebraic argument:
Let $z= e^{ix} = \cos x+i\,\sin x$. Then $$ e^{-ix} = \frac1z = \frac{\bar z}{z\bar z} = \frac{\cos x - i \sin x}{\cos^2 + \sin^2x} = \cos x - i \sin x $$
On the other hand, $$ e^{-ix} = \cos(-x)+i\,\sin(-x) $$
Comparing real and imaginary parts we get $$ \cos(-x) = \cos(x) \qquad \sin(-x) = -\sin(x) $$
lhf
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A question: Is it legitimate to use $\cos{x}-i \sin{x} = \overline{e^{ix}}=e^{-ix}$ instead of using the inversion you did? – user76568 Jan 29 '14 at 15:21
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If more justification is required for this implication, see this answer. – user76568 Jan 29 '14 at 18:54