Addition Formulae
\(\DeclareMathOperator{cosec}{cosec}\)
The Formulae
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\[\sin(A+B)\equiv\sin A\cos B+\cos A\sin B\]
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\[\cos(A+B)\equiv \cos A\cos B-\sin A\sin B\]
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\[\tan(A+B)\equiv \frac{\tan A+\tan B}{1-\tan A\tan B}\]
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\[\sin(A-B)\equiv\sin A\cos B-\cos A\sin B\]
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\[\cos(A-B)\equiv \cos A\cos B+\sin A\sin B\]
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\[\tan(A-B)\equiv \frac{\tan A-\tan B}{1+\tan A\tan B}\]
Show using the formula for \(\sin(A-B)\) that \(\sin(15^{\circ})=\frac{\sqrt{ 6 }-\sqrt{ 2 }}{4}\)
- \(\sin(A-B)=\sin A\cos B-\cos A\sin B\)
- \(\sin(45-30)=\sin45\cos30-\sin30\cos45\)
- \(=\left( \frac{\sqrt{ 2 }}{2}\times \frac{\sqrt{ 3 }}{2} \right)-\left( \frac{1}{2}\times \frac{\sqrt{ 2 }}{2} \right)\)
- \(=\frac{\sqrt{ 6 }}{4}-\frac{\sqrt{ 2 }}{4}\)
- \(=\frac{\sqrt{ 6 }-\sqrt{ 2 }}{4}\)
Show that \(\cos3\theta=4\cos^3\theta-3\cos\theta\)
- \(LHS=\cos3\theta=\cos(\theta+2\theta)\)
- \(=\cos\theta \cos2\theta-\sin\theta \sin2\theta\)
- \(=\cos\theta(2\cos^2\theta-1)-\sin\theta(2\sin\theta \cos\theta)\)
- \(=2\cos^3\theta-\cos\theta-2\sin^2\theta \cos\theta\)
- \(=2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\)
- \(=4\cos^3\theta-3\cos\theta=RHS\)