2

$$\sin^2z+\cos^2z=1\tag a \space \forall z \in \mathbb{C}$$

$(a)\div\sin^2z$

$$1+\cot^2z={\csc^2{z}}\tag b$$

$(a)\div\cos^2z$

$$\tan^2z+1=\sec^2z\tag c$$

$(b)$ and $(c)$ are derived from $(a)$.

But where was $(a)$ originally derived from?

pshmath0
  • 10,565
  • https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi – lab bhattacharjee Dec 21 '17 at 13:00
  • 1
    Pythagoras${}$? – Angina Seng Dec 21 '17 at 13:01
  • $a^2+b^2=c^2$... –  Dec 21 '17 at 13:02
  • Wait... what exactly is your question? You want to know how to show $\sin^2 z + \cos^2 z = 1$? Are the other formulas in any way relevant for the question? Here is a link if that is the only think you ask for https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Pythagorean_identities – Ove Ahlman Dec 21 '17 at 13:05

2 Answers2

3

In the real case, for acute angles use Pythagoras' Theorem. In a right angled triangle,

      /|
    1/ |
    /  | y
   /A__|
      x

with angle $A$, adjacent side $x$, opposite side $y$, and hypotenuse of length $1$, we have the following trig ratios:

$$\cos A=\frac{x}{1}=x\quad\sin A=\frac{y}{1}=y,$$

so we have $$\cos^2A+\sin^2A=x^2+y^2,$$ but $x^2+y^2=1$ by Pythagoras' theorem, so that $$\sin^2A+\cos^2A=1.$$ You can adapt to obtuse angles.

Update

Since it was not initially clear you wanted $z\in\mathbb{C}$, then here is the proof in the complex case:

$$1=e^0=e^{iz-iz}=e^{iz}e^{-iz}=(\cos(z)+i\sin(z))(\cos(-z)+i\sin(-z))\\=(\cos(z)+i\sin(z))(\cos(z)-i\sin(z)),$$ which simplifies to $$\cos^2(z)-i^2\sin^2(z)=\cos^2(z)+\sin^2(z),$$ as @nbubis points out.

pshmath0
  • 10,565
0

If you need to prove this for all $z\in\mathbb{C}$, you can use Euler's identity: $$1 = e^{-iz}e^{iz}= (\cos (-z) + i\sin (-z))(\cos z + i\sin z)=\cos^2 z + \sin^2 z.$$

Nathaniel Bubis
  • 33,018