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en:proof-of-the-law-of-cosines [2017/04/11 11:54] federico [Proof of The Law of Cosines] |
en:proof-of-the-law-of-cosines [2017/07/14 10:36] federico |
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===== Proof ===== | ===== Proof ===== | ||
+ | |||
+ | Divide the triangle into 2 right angle triangles: | ||
+ | |||
+ | {{:en:law-of-cosines2-fs8.png?nolink|}} | ||
+ | |||
+ | Using trigonometry, the sides of the green triangle are | ||
+ | * a | ||
+ | * (b sin θ) | ||
+ | * (c - b cos(θ)) | ||
+ | |||
+ | {{:en:law-of-cosines3-fs8.png?nolink|}} | ||
+ | |||
+ | Using the Pythagorean Theorem: | ||
a<sup>2</sup> = (b sin θ)<sup>2</sup> + (c - b cos(θ))<sup>2</sup> | 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)** |