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 Post subject: trigonometry
PostPosted: Sat, 24 May 2003 16:12:02 UTC 
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Let x=sinA, y=sinB, z=sin(A+B). Express cos(A+B) in the form of the quotient of two polynomials with integral coefficients in x,y, and z.

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PostPosted: Sat, 24 May 2003 18:36:00 UTC 
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sin(A+B)=sinAcosB+cosAsinB\\
=xcosB+ycosA\\
z^2=x^2cos^2A+y^2cos^2B+2xycosAcosB\\
z^2-x^2(1-x^2)-y^2(1-y^2)=2xycosAcosB\\
cosAcosB=\frac{z^2-x^2(1-x^2)-y^2(1-y^2)}{2xy}\\
cos(A+B)=cosAcosB+sinAsinB\\
cos(A+B)=\frac{z^2-x^2(1-x^2)-y^2(1-y^2)+2x^2y^2]}{2xy}

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Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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PostPosted: Sat, 24 May 2003 18:39:18 UTC 
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cool...
I doubt that there's enough space for me to write this answer into the blank, lol~

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 Post subject:
PostPosted: Sat, 24 May 2003 19:32:58 UTC 
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bugzpodder:
It looks nicer if you put a "\" in front of the trig functions, eg. "\sin A".
\sin(A+B)=\sin A\cos B+\cos A\sin B\\
=x\cos B+y\cos A\\
z^2=x^2\cos^2 A+y^2\cos^2 B+2xy\cos A\cos B\\
z^2-x^2(1-x^2)-y^2(1-y^2)=2xy\cos A\cos B\\
\cos A\cos B=\frac{z^2-x^2(1-x^2)-y^2(1-y^2)}{2xy}\\
\cos(A+B)=\cos A\cos B+\sin A\sin B\\
\cos(A+B)=\frac{z^2-x^2(1-x^2)-y^2(1-y^2)+2x^2y^2]}{2xy}


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