The general method I execute in trying to verify trigonometric identities is to

1.) rewrite everything in terms of sin and cos

2.) sum the terms together, i do this for both sides of the identity

3.) now at this point if the identity hasn't been verified, the problem arises where both sides contain a single term, but one side's term is more complicated than the other. Sometimes a greatest common factor is staring me in the face, here it can be easy to cancel down into the simpler side thus the identity is verified.

But, the principle issue I am concerned about is when there is no visible gcf between the numerator and denominator of the more complicated term. In this case,

it becomes a game of manipulating the numerator and denominator with the body of trig identities in a way that would eventually produce an extractable GCF

To illustrate what I mean if this is vague, here is a simple identity:


the LHS is clearly more complicated than the RHS, so I will try to reduce the LHS into the RHS. But there is no common factor at all between the numerator and denominator, so I must use general algebraic operations, and trig identities to manipulate them until eventually a common factor appears.

Here it is done by spotting a difference of two squares and rewriting the denominator:

$$\frac{(1-\sin^2(x))(1+\sin^2(x)}{(1-\sin^2(x))\cos^2(x)} =\frac{1+\sin^2(x)}{\cos^2(x)} $$

But is there a general procedure for doing this? or is it all about "experience" "intuition" and "doing lots of problems?"

So far, the progress I've made on this problem is:

  1. Since I wrote the entire equation in terms of sin and cos, only the identities that exclusively mention sin and cos and their natural powers will be relevant.

  2. Analysing the differences between the corresponding numerators and corresponding denominators

a difference of two powers, or a sign change suggests the usage of the pythagorean identity

  1. Difference of two squares and three squares formulae can be helpful sometimes

is there anything further I can do to reliably and efficiently find these hidden common factors between the numerators and denominators of trigonometric fractions?

  • 2
    $\begingroup$ Cases where everything is expressed in terms of $\sin^2 x, \cos^2 x$ are particularly easy to solve because they reduce to finding the $\gcd$ of two polynomials. A fraction like $\frac{P(\sin^2x,\cos^2x)}{Q(\sin^2x,\cos^2x)}$ can be simplified iff $\deg \gcd\big(P(t, 1-t),Q(t, 1-t))\big) \gt 0$. $\endgroup$
    – dxiv
    Apr 22, 2023 at 23:31
  • 1
    $\begingroup$ You can just cross-multiply and express through a single angle; you don't need the gcd. But the question of finding gcds in the trigonometric polynomial ring is an interesting one. $\endgroup$ Apr 22, 2023 at 23:46
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    $\begingroup$ Actually, it's not clear whether a gcd even make sense for trigonometric polynomials. For instance, you have $\dfrac{\sin x}{1-\cos x} = \dfrac{1+\cos x}{\sin x}$, which you can easily prove by cross-multiplying, but you cannot cancel anything on either side (unless you pass to $x/2$). $\endgroup$ Apr 22, 2023 at 23:49
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    $\begingroup$ @Maalik In this case, yes. For your example you would have $\frac{1-t^2}{(1-t)^2}$ and $\gcd(1-t^2, (1-t)^2) = 1-t$ which means that a factor of $1-\sin^2x=\cos^2x$ cancels out. $\endgroup$
    – dxiv
    Apr 22, 2023 at 23:55
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    $\begingroup$ Maybe it would be helpful to use the identities $\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$. Just my two cents. $\endgroup$
    – IAAW
    Apr 23, 2023 at 1:22


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