There are certain identities that help us to determine the values of trigonometric functions at $\dfrac{\pi}{2}+x \text{, } \pi-x$ etc. given the values of $\sin x, \cos x$.

Now, when we prove such identities, we usually take the value of $x$ to be in the interval $\Big (0, \dfrac{\pi}{2} \Big )$. Isn't it necessary to prove the identities by taking the value of $x$ in all $4$ quadrants individually and then arriving at the outcome? If not, then why not?

Pardon me if this sounds silly.



If the basis of your proofs are from Euler's identity $e^{i \theta} = \cos \theta + i \sin \theta$ then the quadrant becomes irrelevant.

Also consider the identities $\cos \theta = \dfrac {e^{i \theta} + e^{-i \theta} } 2$ and $\sin \theta = \dfrac {e^{i \theta} - e^{-i \theta} } 2$.

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  • $\begingroup$ The concept of complex numbers has not been introduced to me yet. In my Mathematics textbook, they have been proved using a unit circle and finding the coordinates of the points where the allied angles intersect with the unit circle. Shall I elaborate? $\endgroup$ – Rajdeep Sindhu May 27 at 15:00
  • $\begingroup$ No, I understand the form. My answer would be: yes, I would apply a geometrical argument to all 4 quadrants. A proof that did not would probably not satisfy me if I were doing it that way. BTW not a silly question at all. $\endgroup$ – Prime Mover May 27 at 15:02
  • $\begingroup$ It doesn't satisfy me too! Thanks $\endgroup$ – Rajdeep Sindhu May 27 at 15:04
  • $\begingroup$ By the way, the identities that you mention in your answer, they are related to complex numbers, right? $\endgroup$ – Rajdeep Sindhu May 27 at 15:04
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    $\begingroup$ You have no idea about how much troubled and confused I was due to that algorithm. While I find some algorithms helpful, I hate all mnemonics like the one for remembering the signs for trigonometric functions in various quadrant All School To College. I wish there was a 'follow' option on this website. Thanks, again! PS : The 'algorithm' is useful for making a computer program though $\endgroup$ – Rajdeep Sindhu May 27 at 15:36

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