Four Muffins
New member
- Joined
- Jun 29, 2022
- Messages
- 25
Hello again. I got stuck for a while on proving [imath]cos(\alpha-\beta) = cos \alpha cos\beta+sin\alpha sin\beta[/imath]. I did solve it, but only by doing something that seems obviously wrong.
With the Cosine Law, [imath]c^2=a^2+b^2-2ab*cos(\theta)[/imath], I normally set [imath]c[/imath] to be the hypotenuse. The triangle in the question has the shortest side of an isosceles triangle as [imath]c[/imath], so I set one of the equal sides to be the pseudo-hypotenuse [imath]b[/imath] and solved for [imath]c[/imath], getting [imath]c^2=b^2-a^2+2ab*cos(\alpha -\beta)[/imath], which did not let me prove the formula.
Does the Cosine Law allow for solving any one unknown side without algebraic manipulation, or am I missing something about what [imath]c[/imath] is in this triangle?
Correct, then incorrect working.
With the Cosine Law, [imath]c^2=a^2+b^2-2ab*cos(\theta)[/imath], I normally set [imath]c[/imath] to be the hypotenuse. The triangle in the question has the shortest side of an isosceles triangle as [imath]c[/imath], so I set one of the equal sides to be the pseudo-hypotenuse [imath]b[/imath] and solved for [imath]c[/imath], getting [imath]c^2=b^2-a^2+2ab*cos(\alpha -\beta)[/imath], which did not let me prove the formula.
Does the Cosine Law allow for solving any one unknown side without algebraic manipulation, or am I missing something about what [imath]c[/imath] is in this triangle?
Correct, then incorrect working.