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$$\cos(\theta) = \sin \left(\tfrac{\pi}{2} - \theta\right)$$ $$\sin(\theta) = \cos \left(\tfrac{\pi}{2} - \theta\right)$$

Both are the same entity. But is sine the copy of cosine, or is cosine the copy of sine?

If you really don't see any difference between these two functions, my question could be rephrased as such:

If for the rest of your life you had to only use one of the two following pairs of functions:

  • $\cos(\theta)$ and $\cos \left(\frac{\pi}{2} - \theta\right)$

  • $\sin(\theta)$ and $\sin \left(\frac{\pi}{2} - \theta\right)$

...which pair would you choose?

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closed as off-topic by Sanath Devalapurkar, O.L., Sujaan Kunalan, user7530, Qiaochu Yuan May 17 at 5:10

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I would choose $\sin(\theta)$. My thought is its sign per quadrant. $\{+,+,-,-\}$ whereas cosine is $\{+,-,-,+\}$ Not really mathematical decision, but you asked... –  Iceman May 16 at 23:38
3  
Sorry, but this question is senseless. –  A little lime May 16 at 23:39
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The most fundamental trigonometric function is the exponent. –  Artem May 16 at 23:40
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I like $\sin$ because $\sin0=0$ is easier to remember. And sine is easier to say than cosine. –  user137794 May 16 at 23:46
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The exponential function is fundamental. The trigonometric functions $\sin$, $\cos$, $\sinh$ and $\cosh$ are just piggybacking. –  mjqxxxx May 17 at 0:03

4 Answers 4

up vote 4 down vote accepted

The cosine is nicer because it corresponds to the real part of the exponential of imaginary numbers and its power series is invertible.

The sine is nicer because it is an increasing function for angles in the first quadrant.

For substitution purposes in integrals, you should not neglect the option of choosing $\tan \frac{\theta}2$ as basic.

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Good arguments! I didn't thought about that. I would have think sine is more fundamental than cosine because we can express cosine in term of sine in a simpler manner than we can express sine in term of cosine: $cos(\theta) = sin(\theta + \frac{\pi}{2})$ is a little nicer and simpler than $sin(\theta) = cos(\theta - \frac{\pi}{2})$ Also, the derivative of sine is cosine, but the derivative of cosine is minus sine which is a little less nice. –  Omega Force May 16 at 23:58
    
But if you choose $\cos(\theta)$, you can never write or speak of $\sin(\theta)$, so how would you know these facts about the sine function? –  David K May 17 at 1:44

The premise of the question -- that there is any reason to choose one over the other -- is invalid.

If I had to do without one of them, then fine: I would do without both. For, much of what can be done with sines and cosines (e.g. Fourier series) is more elegant with $e^{iz}$.

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Agreed. (Though I am biased towards $\cos(x)$ because of the Discrete Cosine Transform) –  Navin May 17 at 4:55

While I tend to agree with the obvious response that there is no reason to favor one over the other, I think it is interesting to note that there is a sense in which $\cos(\theta)$ can be considered the 'preferred' function: namely, when we think of $\sin(\theta)$ and $\cos(\theta)$ as being defined by functional equations.

As it turns out, the functional equation $ C(x-y) = C(x) C(y) + S(x) S(y)$ for all $x, y \in \mathbb{R}$, where $S: \mathbb{R} \to \mathbb{R}$ and $C: \mathbb{R} \to \mathbb{R}$, along with a given value of each function and an assertion about their positivity, uniquely define the familiar sine and cosine functions.

Meanwhile, the functional equation $ S(x + y) = S(x) C(y) + C(x) S(y)$ for all $x, y \in \mathbb{R}$ does not yield unique solutions when it is subject to the same conditions. Instead, we obtain the functions $S(x) = a^{x-\lambda} \sin (x)$ and $C(x) = a^x \cos (x)$, where $\lambda$ is positive and depends on the conditions I mentioned, and $a$ is any positive real number.

The details and some discussion can be found in Efthimiou's Introduction to Functional Equations.

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They are tightly coupled: $$ \sin'= \cos, \qquad \cos'=-\sin $$ Not surprisingly, this translates to $$ \pmatrix { \cos(x) \\ \sin(x) }'= \pmatrix{0 & -1 \\ 1 & \hphantom{-}0 } \pmatrix { \cos(x) \\ \sin(x) } $$ and $$ \pmatrix{0 & -1 \\ 1 & \hphantom{-}0 } ^ 2 = \pmatrix{1 & 0 \\ 0 & 1} $$ which is the expression of $(e^{ix})'= i e^{ix}$.

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