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--- en:proof-of-the-law-of-cosines [2017/04/11 10:46]
federico
+++ en:proof-of-the-law-of-cosines [2017/07/14 10:36]
federico
@@ Line 1: / Line 1: @@
 ====== 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|}}
+**Law of Cosines**: a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2bccos(θ)
 ===== 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>
@@ Line 9: / Line 24: @@
 = 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> + c<sup>2</sup> - 2cbcos(θ)
+**a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2cbcos(θ)**