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en:proof-of-the-law-of-cosines [2017/04/11 10:44] federico created |
en:proof-of-the-law-of-cosines [2017/07/14 10:47] (current) federico [Proof] |
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| ====== Proof of The Law of Cosines ====== | ====== Proof of The Law of Cosines ====== | ||
| - | **Law of Cosines**: a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2 bc cos Θ | + | {{:en:law-of-cosines1-fs8.png?nolink|}} |
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| + | **Law of Cosines**: a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2bccos(θ) | ||
| ===== Proof ===== | ===== Proof ===== | ||
| - | a<sup>2</sup> = (b sin Θ)<sup>2</sup> + (c - b cos(Θ))<sup>2</sup> | + | Divide the triangle into 2 right angle triangles: |
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| + | {{:en:law-of-cosines2-fs8.png?nolink|}} | ||
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| + | Using trigonometry, the sides of the green triangle are | ||
| + | * Longest side: **a** | ||
| + | * Side on the left: **b . sin θ** ; because the sine of θ is opposite divided the hypotenuse, and the opposite is the length of this side. Then, the length of this side is the hypotenuse multiplied by the sine of θ | ||
| + | * Side on the right: ***c - b cos(θ)** ; we must subtract from the length "c", the part of c that is occupied by the red triangle. We know that, in the red triangle, cos θ is the adjacent side divided the hypotenuse. The adjacent is the "base" of the red triangle, and it is cos θ multiplied by the hypotenuse. The hypotenuse is b. Then, the base of the red triangle is (cos θ)b. So, the length of the "base" of the green triangle is c - b.cos θ | ||
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| + | {{:en:law-of-cosines3-fs8.png?nolink|}} | ||
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| + | Using the Pythagorean Theorem: | ||
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| + | a<sup>2</sup> = (b sin θ)<sup>2</sup> + (c - b cos(θ))<sup>2</sup> | ||
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| + | = b<sup>2</sup> sin <sup>2</sup> θ + c<sup>2</sup> - 2cbcos(θ) + b <sup>2</sup> cos <sup>2</sup> θ | ||
| - | = b<sup>2</sup> sin <sup>2</sup> Θ + c<sup>2</sup> - 2cbcos(Θ) + b <sup>2</sup> cos <sup>2</sup> Θ | + | = b<sup>2</sup> (sin <sup>2</sup> θ + cos <sup>2</sup> θ) + c<sup>2</sup> - 2cbcos(θ) |
| - | = b<sup>2</sup> (sin <sup>2</sup> Θ + cos <sup>2</sup> Θ + c<sup>2</sup> - 2cbcos(Θ) | + | Knowing that (sin <sup>2</sup> θ + cos <sup>2</sup> θ = 1) ((see https://en.wikibooks.org/wiki/Trigonometry/Sine_Squared_plus_Cosine_Squared)) |
| - | = b<sup>2</sup> * 1 + c<sup>2</sup> - 2cbcos(Θ) | + | = b<sup>2</sup> * 1 + c<sup>2</sup> - 2cbcos(θ) |
| - | = b<sup>2</sup> + c<sup>2</sup> - 2cbcos(Θ) | + | **a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2cbcos(θ)** |