# Help me proof the identity $\sin^2\alpha+\cos^2\alpha=1$

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

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This follows at once from the (extension of the) definition of the trigonometric functions on the unit circle $\,x^2+y^2=1\,$ . Almost any basic algebra book has this stuff. – DonAntonio Sep 27 '12 at 14:28
One typically constructs a right-angled triangle (with hypotenuse 1) and derives from the Pythagoran theorem. Then you need to prove that this relationship is independent of the size fo hypotenuse (i.e. for similar triangles it would hold). – gt6989b Sep 27 '12 at 14:29
This is equivalent to the Pythagorean theorem: en.wikipedia.org/wiki/… – Dennis Gulko Sep 27 '12 at 14:37

$$\sin\alpha=\frac{a}{c}\Rightarrow\sin^2 \alpha=\frac{a^2}{c^2}$$

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

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

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Missing squares. And, you haven't told us anything about $a$, $b$, or $c$, so how can we understand why $a^2 + b^2 = c^2$ is true? – Graphth Sep 27 '12 at 14:51
Everybody knows that $a^2+b^2=c^2$ :-) – draks ... Sep 27 '12 at 14:58
If so, then why should one prove it? – awllower Sep 27 '12 at 15:06
@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..." – Pedro Tamaroff Sep 27 '12 at 15:26

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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. – DonAntonio Sep 27 '12 at 17:17

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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I begin to wonder about the possibility that you have read the books by Landau...I kid here: no intention to be rude. – awllower Sep 27 '12 at 15:08
@awllower I do. I really enjoy his expositions. This approach is also used by Apostol. – Pedro Tamaroff Sep 27 '12 at 15:09
I am glad to find someone who enjoys his books, too!Per chance you like as well the theory of numbers? – awllower Sep 27 '12 at 15:10
@awllower Yes, I studied from that one too. =) – Pedro Tamaroff Sep 27 '12 at 15:11
@awllower Well, I think more people should read Landau's exposition. He's very slick in his proofs! – Pedro Tamaroff Sep 27 '12 at 15:17

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"?? – DonAntonio Sep 27 '12 at 17:18