Prove the identity $\sin^2\alpha+\cos^2\alpha=1$. Thanks
Help me proof the identity $\sin^2\alpha+\cos^2\alpha=1$
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0This 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. – 2012-09-27
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0One 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). – 2012-09-27
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0This is equivalent to the Pythagorean theorem: http://en.wikipedia.org/wiki/Pythagorean_theorem#Pythagorean_trigonometric_identity – 2012-09-27
4 Answers
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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0What is "perp"?? – 2012-09-27
$\hskip2.3in$
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0Very 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
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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0I begin to wonder about the possibility that you have read the books by Landau...I kid here: no intention to be rude. – 2012-09-27
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0@awllower I do. I really enjoy his expositions. This approach is also used by Apostol. – 2012-09-27
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0I am glad to find someone who enjoys his books, too!Per chance you like as well the theory of numbers? – 2012-09-27
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0@awllower Yes, I studied from that one too. =) – 2012-09-27
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0Haha, a real coincidence! – 2012-09-27
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1@awllower Well, I think more people should read Landau's exposition. He's very slick in his proofs! – 2012-09-27
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0I wonder how "basic" the sum/difference of angles formula for the cosine is *without* first proving the trigonometric Pythagorean theorem... – 2012-09-27
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0@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
$$\sin\alpha=\frac{a}{c}\Rightarrow\sin^2 \alpha=\frac{a^2}{c^2}$$
$$\cos\alpha=\frac{b}{c}\Rightarrow\cos^2\alpha=\frac{b^2}{c^2}$$
$$\sin^2\alpha+\cos^2\alpha=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1$$
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0Missing 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? – 2012-09-27
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2Everybody knows that $a^2+b^2=c^2$ :-) – 2012-09-27
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0If so, then why should one prove it? – 2012-09-27
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2@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