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Prove the identity $\sin^2\alpha+\cos^2\alpha=1$. Thanks

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    This is equivalent to the Pythagorean theorem: http://en.wikipedia.org/wiki/Pythagorean_theorem#Pythagorean_trigonometric_identity2012-09-27

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Take a right angled triangle with one angle $\alpha$, then,

Let length of the side opposite to the angle $\alpha$ be $x$

and length of the second side other than Hypotenuse be $y$

$\sin\alpha=\frac{x}{Hypotenuse}$

and $\cos\alpha=\frac{y}{Hypotenuse}$

Then, $\sin^2\alpha+\cos^2\alpha=\frac{(x)^2+(y)^2}{(Hypotenuse)^2}=\frac{(Hypotenuse)^2}{(Hypotenuse)^2}=1$

Here, i have used Pythagoras theorem, $(x)^2+(y)^2=(Hypotenuse)^2$

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    What is "perp"??2012-09-27
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    Very nice, but it seems to be a long shot to expect this to be understood by someone making as basic a question as the OP.2012-09-27
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We have that

$\begin{eqnarray*}1=\cos(0)=\cos(x-x)&=&\cos x\cos x+\sin x \sin x\\&=&\cos^2x+\sin^2x\end{eqnarray*}$

Given the fundamental identity: $\cos(x-y)=\cos x\cos y+\sin x\sin y$

and the fundamental value of the cosine at $0$.

$1=\cos(0)$

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    @DonAntonio I said **fundamental**, not basic, whatever that means to you. What does **fundamental** mean? If two functions $\sin$ and $\cos$ satisfy $4$ basic properties (the angle difference being one), then they satisfy all other properties (such as the Pythagorean identity) which we "want" them to, regardless of how we define them (circular, power series, integrals). I can expand on this if you're interested.2012-09-27
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$\sin\alpha=\frac{a}{c}\Rightarrow\sin^2 \alpha=\frac{a^2}{c^2}$

$\cos\alpha=\frac{b}{c}\Rightarrow\cos^2\alpha=\frac{b^2}{c^2}$

$\sin^2\alpha+\cos^2\alpha=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1$

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    @draks One should be more detailed. For example, one can add: "Let $a$,$b$,$c$ be the sides of$a$right triangle, $c$ being the hypothenuse. By the Pythagorean theorem, $a^2+b^2=c^2$. But from the circular definition of the sine and cosine..."2012-09-27