BGN-2.1.2

Geometry 几何

theorems 定律

1)In any triangle ABCABC, AB<AC+BCAB\lt AC+BC.
1)对于任意三角形 ABCABCAB<AC+BCAB\lt AC+BC
2)For any right-angled triangle ABCABC with side aa as hypotenuse, a2=b2+c2a^2=b^2+c^2.
2)对于任意以 aa 为斜边的直角三角形 ABCABCa2=b2+c2a^2=b^2+c^2
3)For any triangle ABCABC, sinAa=sinBb=sinCc\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}.
3)对于任意三角形 ABCABCsinAa=sinBb=sinCc\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}
4)For any triangle ABCABC, a2=b2+c22bccosAa^2=b^2+c^2-2bc\cos A.
4)对于任意三角形 ABCABCa2=b2+c22bccosAa^2=b^2+c^2-2bc\cos A

proof 证明

Please try to prove 2) and 3) yourself using size relationships.
请自行尝试用面积关系证明定律 2) 和 3) 。

Trigonometry 三角学

theorems 定律

1) sin2θ+cos2θ=1(sin2θ=(sinθ)2)\sin ^2 \theta + \cos ^2 \theta = 1 {\kern 5em} ( \sin ^2 \theta = ( \sin \theta )^2)
2) sin(A±B)=sinAcosB±cosAsinB\sin (A\pm B) =\sin A \cos B \pm \cos A\sin B
3) cos(A±B)=cosAcosBsinAsinB\cos (A\pm B)=\cos A\cos B \mp \sin A\sin B
4) tan(A±B)=tanA±tanB1tanAtanB\tan (A\pm B)=\dfrac{\tan A \pm \tan B}{1\mp \tan A\tan B}
5) sin2α=2sinαcosα\sin 2\alpha =2\sin \alpha \cos \alpha
6) cos2α=cos2αsin2α=2cos2α1=12sin2α\cos 2\alpha = \cos ^2\alpha -\sin^2\alpha =2\cos ^2\alpha -1=1-2\sin ^2 \alpha
7) tan2α=2tanα1tan2α\tan 2\alpha = \dfrac{2\tan \alpha}{1-\tan ^2\alpha}
8) sin2A=1cos2A2\sin ^2 A=\dfrac{1-\cos 2A}{2}
9) cos2A=1+cos2A2\cos ^2 A=\dfrac{1+\cos 2A}{2}
10) tan2A=1cos2A1+cos2A\tan ^2 A=\dfrac{1-\cos 2A}{1+\cos 2A}
11) For a unit circle on a coordinate grid with its center at the origin, point P(cosθ,sinθ)P(\cos \theta ,\sin \theta ) is on the circle, and line OP\overline{OP} and the x-axis form an angle of θ\theta.
11)对于一个表示在坐标轴上且圆心在原点的单位圆,点 P(cosθ,sinθ)P(\cos \theta ,\sin \theta ) 一定在该圆上,而且直线 OP\overline{OP} 和x-轴的夹角为 θ\theta

proof 证明

11) will be used but not proved 11) 会被用来证明,但它本身不会被证明
Try to prove 1),5),6),7).
尝试证明 1),5),6),7).

Other proofs 其他证明:

Note by Jeff Giff
1 month, 1 week ago

No vote yet
1 vote

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Comments

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@Páll Márton, I’ve been thinking of a proof for 8,9,10, but I have no idea. Can you help me?

Jeff Giff - 1 month ago

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Yeah. I can. \hspace{50px}

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Never mind :) I finally found the proof online

Jeff Giff - 1 month ago

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LOL Your first(second) line is wrong

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oops... third

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or fourth with the title

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Where?

Jeff Giff - 1 month ago

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@Jeff Giff Wait...whoa! Thanks!

Jeff Giff - 1 month ago

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@Jeff Giff ABC, AB\gt AC+BC

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P.S. I added one proof to the chain

Jeff Giff - 1 month ago

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In China the correct order is 2,3,1,5,6,7??? lol

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@Páll Márton (no activity) Umm... actually there are two parts, and the numbers refer to the theorem in the part :)

Jeff Giff - 1 month ago

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