Using the diagram above, we can show two angles, [math]\alpha[/math] and [math]\beta[/math] in standard position. Let [math]T[/math] and [math]R[/math] be the points where their terminal sides meet the unit circle. [i]Show Alpha and Beta using the checkboxes.[br][math]T\left(\cos\alpha,\sin\alpha\right)[/math][br][/i][math]R\left(\cos\beta,\sin\beta\right)[/math][br][br][math]m\angle TOR=\alpha-\beta[/math] [i]Show the difference[/i][br]Now, we can put this angle into standard position, by rotating the triangle.[br][br][i]Check "Rotation" and use the slider to rotate the triangle into standard position. Check "Rotated Labels"[/i][br][size=100]If [math]S[/math] is the point where the terminal side of [math]\alpha-\beta[/math], now in standard position, meets the unit circle, then it has coordinates:[br][math]S\left(\cos\left(\alpha-\beta\right),\sin\left(\alpha-\beta\right)\right)[/math][math]Q\left(1,0\right)[/math][/size]
Why is [math]\triangle TOR\cong\triangle SOQ[/math]?
Answer 1. One is a rotation of the other and rotations take figures to congruent figures.[br]Answer 2. By SAS, since [math]\overline{OR}\cong\overline{OT}\cong\overline{OS}\cong\overline{OQ}[/math] as they are all radii of the same circle. Clearly [math]\angle TOR\cong\angle SOQ[/math] since they both have a measure of [math]\alpha-\beta[/math].
[math]TR=SQ[/math][br]Use the coordinates of [math]T,R,S,Q[/math] and the distance formula to rewrite this equation. Square both sides. Expand everything and simplify. Finally solve for [math]\cos\left(\alpha-\beta\right)[/math].
[math]\sqrt{\left(\cos\alpha-\cos\beta\right)^2+\left(\sin\alpha-\sin\beta\right)^2}=\sqrt{\left(\cos\left(\alpha-\beta\right)-1\right)^2+\left(\sin\left(\alpha-\beta\right)-0\right)^2}[/math][br][math]\left(\cos\alpha-\cos\beta\right)^2+\left(\sin\alpha-\sin\beta\right)^2=\left(\cos\left(\alpha-\beta\right)-1\right)^2+\sin^2\left(\alpha-\beta\right)[/math][br][math]\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta=\cos^2\left(\alpha-\beta\right)-2\cos\left(\alpha-\beta\right)+1+\sin^2\left(\alpha-\beta\right)[/math][br][br]Notice: [math]\cos^2\alpha+\sin^2\alpha=1[/math], [math]\cos^2\beta+\sin^2\beta=1[/math] and [math]\cos^2\left(\alpha-\beta\right)+\sin^2\left(\alpha-\beta\right)=1[/math][br][br][math]1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=2-2\cos\left(\alpha-\beta\right)[/math][br][math]-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=-2\cos\left(\alpha-\beta\right)[/math][br][br][math]2\cos\left(\alpha-\beta\right)=2\cos\alpha\cos\beta+2\sin\alpha\sin\beta[/math][br][br][math]\cos\left(\alpha-\beta\right)=\cos\alpha\cos\beta+\sin\alpha\sin\beta[/math]
Now apply this formula using the angles [math]\alpha[/math] and [math]-\beta[/math]. Use even and odd properties to simplify so everything is expressed in terms of functions of [math]\alpha[/math] and [math]\beta[/math].[br]
[math]\cos\left(\alpha-\left(-\beta\right)\right)=\cos\alpha\cos\left(-\beta\right)+\sin\alpha\sin\left(-\beta\right)[/math][br]Because cosine is an even function, [math]\cos\left(-\beta\right)=\cos\beta[/math], [br]Sine is an odd function so [math]\sin\left(-\beta\right)=-\sin\beta[/math], which leaves us with:[br][math]\cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta+\sin\alpha\left(-\sin\beta\right)[/math][br][math]\cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta[/math][br][br]