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What are the general procedures for simplifying a trigonometric expression using Euler's formula? As an example, this is how we can simplify the following trigonometric expression:

$$\sin{x}\cos{x}$$

$$\sin{x}\cos{x} = \dfrac{e^{ix}-e^{-ix}}{2i} \times \dfrac{e^{ix}+e^{-ix}}{2}$$

$$ = \dfrac{(e^{ix}-e^{-ix})(e^{ix}+e^{-ix})}{4i}$$ $$ = \dfrac{(e^{2ix}-e^{-2ix})}{4i}$$ $$ = \dfrac{1}{2} \times \dfrac{(e^{i(2x)}-e^{-i(2x)})}{2i}$$ $$ = \dfrac{1}{2} \times \sin(2x)$$

However, how would you approach other expressions? What general steps can you take to simplify any trigonometric expression, such as the following:

$$ \tan^{-1}(\dfrac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}})$$ Where $\dfrac{-\pi}{4} < x < \dfrac{\pi}{4}$

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    $\begingroup$ Note that since $\sin(2x) = 2\sin x\cos x$, then $\sin x\cos x = \frac{1}{2}\times\sin(2x)$, i.e., it's not $\frac{1}{2}\times\cos(2x)$ as you wrote. $\endgroup$
    – John Omielan
    Commented Jul 30, 2023 at 2:12
  • $\begingroup$ For this particular identity it's also efficient to write $x = u + \frac\pi4$, expand using angle sum identities, and simplify. $\endgroup$
    – Travis Willse
    Commented Jul 30, 2023 at 2:49
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    $\begingroup$ Did you think that $\sin x$ is equal to $\dfrac{e^{ix}-e^{-ix}}{2}$? It isn't. You seem to have your formulas mixed up. $\endgroup$
    – David K
    Commented Jul 30, 2023 at 2:55
  • $\begingroup$ Sorry about the error; it should hopefully be fine now. $\endgroup$
    – Shane Sarosh ℼℼℼ
    Commented Jul 30, 2023 at 11:40

1 Answer 1

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Replace the sines and cosines by the $e^{ix}$ expressions, and simplify. That'll make the argument to arctan a lot simpler. In fact, if you write $$ u = \tan^{-1}(\dfrac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}}) $$ then you can rewrite that as $$ \tan u = \dfrac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}} $$ The stuff on the right simplifies to some exponential involving $x$; the stuff on the left similarly simplifies to something involving $u$. Set these equal, and see whether you can find a relationship between $u$ and $x$ as a result.

You'll notice I haven't done the work for you --- that's because you asked "what general steps can be taken," and I've answered that.

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    $\begingroup$ One more hint: at some point you're going to end up with $e^{ix}$ all over the place, and $e^{-ix}$ as well. If you use the letter $q$ for the first, then you can use $1/q$ for the second; you can do something similar with $e^{iu}$ (call it $r$, say). Then you end up with a polynomial relation between $q$ and $r$ which you can simplify before substituting back. $\endgroup$
    – John Hughes
    Commented Jul 30, 2023 at 2:54

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