Simplifying \(a \cos(x)^+_{-} b \sin(x)\)
\(\DeclareMathOperator{cosec}{cosec}\)
R formulae
- For positive values of \(a\) and \(b\):
- \(a \sin x^+_{-}b\cos x\) can be expressed in the form \(R\sin(x^+_{-}\alpha)\)
- \(a \cos x^+_{-}b\sin x\) can be expressed in the form \(R\cos(x^-_{+}\alpha)\)
- \(R>0\) and \(0<\alpha<90^{\circ}\) or \(\frac{\pi}{2}\)
- Where \(R\cos\alpha=a\), \(R\sin\alpha=b\), \(R=\sqrt{ a^2+b^2 }\)
Express \(2\cos\theta+5\sin\theta\) in the form \(R\cos(\theta-\alpha)\)
- \(2\cos\theta+5\sin\theta\equiv R\cos\theta \sin\alpha+R\sin\theta \cos\alpha\)
- \(R\cos\alpha=2\) and \(R\sin\alpha=5\)
- \(\frac{R\sin\alpha}{R\cos\alpha}=\frac{5}{2}\)
- \(\tan\alpha=\frac{5}{2}\)
- \(\alpha=68.2^{\circ}\)
- \(R=\sqrt{ a^2+b^2 }=\sqrt{ 29 }\)
- \(\equiv \sqrt{ 29 }\cos(\theta-68.2)\)
Show that you can express \(3\sin x+4\cos x\) in the form \(R\sin(x+\alpha)\) and in the form \(R\cos(x-\alpha)\)
- Let \(3\sin x+4\cos x\equiv R\sin x\cos\alpha+R\cos x\sin\alpha\)
- So \(R\cos\alpha=3\) and \(R\sin\alpha=4\)
- \(\tan\alpha=\frac{4}{3}\)
- \(\alpha=53.1^{\circ}\)
- \(R=\sqrt{ 3^2+4^2 }=5\)
- Therefore: \(3\sin x+4\cos x=5\sin(x+53.1^{\circ})\)
- \(3\sin x+4\cos x\equiv R\cos x\cos\alpha+\sin x\sin\alpha\)
- \(R\cos\beta=4\)
- \(R\sin\beta=3\)
- \(\frac{R\sin\beta}{R\cos\beta}=\frac{3}{4}\)
- \(\tan\beta=\frac{3}{4}\)
- \(\beta=36.9^{\circ}\)
- \(R=\sqrt{ 3^2+4^2 }\)
- \(3\sin x+4\cos x=5\cos(x-36.9^{\circ})\)
\(5\sin\theta+12\cos\theta\equiv R\sin(\theta+\alpha)\)
- \(5\sin\theta+12\cos\theta\equiv R\sin x\cos\alpha+R\cos x\sin\alpha\)
- \(R\sin\alpha=5\) and \(R\sin\alpha=12\)
- \(\tan\alpha=\frac{12}{5}\)
- \(R=\sqrt{ 12^2+5^2 }=13\)