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I know that sine and cosine can be rewritten in terms of the real and complex parts of the exponential function as a result of Euler's formula.

My question is, can every trigonometric expression be written in terms of elementary trigonometric functions ($\sin$, $\cos$)? If not, why couldn't they be?

I would think that they could, although I understand that sometimes it may be prohibitive to do so since most trig identities can be derived from Euler's formula. Are there any cases when a trig expression absolutely cannot be written in terms of the elementary functions?

The only potential counterexamples I could think of would include some non trigonometric terms or factors.

I know that hyperbolic sine and cosine can be rewritten in terms of sine and cosine in the complex plane.

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    $\begingroup$ Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos. $\endgroup$
    – Eleven-Eleven
    Jan 20, 2019 at 17:28
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    $\begingroup$ If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$ $\endgroup$
    – Andrei
    Jan 20, 2019 at 17:28
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    $\begingroup$ It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that? $\endgroup$
    – David K
    Jan 20, 2019 at 17:40
  • $\begingroup$ By the way, you can also express cosine in terms of sine (or of course the other way around), as in $\cos(x) = \sin(x+\frac{\pi}{2})$, or $\sin(x) =\begin{cases}\sqrt{1-\cos^2(x)}\text{ if } x \in [2k\pi, (2k+1)\pi]\\-\sqrt{1-\cos^2(x)}\text{ if } x \in [(2k+1)\pi, (2k+2)\pi] \end{cases}, k \in \mathbb Z$ -- if that matches your definition of "in terms of". From a numerical viewpoint, as soon as you have one of the trigonometric functions and a good enough approximation of $\pi$, you have them all. $\endgroup$
    – Torsten Schoeneberg
    Jul 13, 2019 at 17:14

1 Answer 1

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$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$

$$\sec\theta=\frac{1}{\cos\theta}$$

$$\csc\theta=\frac{1}{\sin\theta}$$

and since in the complex plane, we have

$$\begin{align} \cosh\theta&=\phantom{-i}\cos{i\theta} \\ \sinh\theta&=-i\sin{i\theta} \\ \tanh\theta&=-i\tan{i\theta} \\ \coth\theta&=\phantom{-}i\cot{i\theta} \\ \operatorname{sech}\theta&=\phantom{-i}\sec{i\theta} \\ \operatorname{csch}\theta&=\phantom{-}i\csc{i\theta} \end{align}$$

And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.

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  • $\begingroup$ Not sure why sech and csch don't render right? $\endgroup$
    – Eleven-Eleven
    Jan 20, 2019 at 17:39

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