Double-angle formulae
\(\DeclareMathOperator{cosec}{cosec}\)
Double angle formulae derivation for sin
- If angles are the same:
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\[\sin(A+B)=\sin A\cos B+\cos A\sin B\]
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\[A=B\]
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\[\sin2A=\sin A\cos A+\cos A\sin A\]
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\[\sin2A=2\sin A\cos A\]
Double angle formulae derivation for cos
- If angles are the same:
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\[\cos(A+B)=\cos A\cos B-\sin A\sin B\]
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\[A=B\]
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\[\cos2A=\cos A\cos A-\sin A\sin A\]
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\[\cos2A=\cos^2A-\sin^2A\]
Use \(\sin^2\theta=1-\cos^2\theta\) to find another formulae for cos in terms of sin
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\[\cos^2A=1-\sin^2A\]
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\[\cos2A=1-\sin^2A-\sin^2A\]
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\[\cos2A=1-2\sin^2A\]
Use \(\sin^2\theta=1-\cos^2\theta\) to find another formulae for cos in terms of cos
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\[\sin^2A=1-\cos^2A\]
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\[\cos2A=\cos^2A-(1-\cos^2A)\]
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\[\cos2A=2\cos^2A-1\]
Double angle formulae derivation for tan
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\[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]
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\[A=B\]
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\[\tan2A=\frac{\tan A+\tan A}{1-\tan A\tan A}\]
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\[\tan2A=\frac{2\tan A}{1-\tan^2A}\]
Solve the equation: \(\cos2x-5\cos x=2\)
- \(2\cos^2x-1-5\cos x=2\)
- \(2\cos^2x-5\cos x-3=0\)
- \((2\cos x+1)(\cos x-3)=0\)
- \(\cos x=-\frac{1}{2}\)
- \(\cos x=3\)
- \(x=\frac{2\pi}{3}\)
- \(x=\frac{4\pi}{3}\)
Prove that \(2\cot \frac{x}{2}\left( 1-\cos^2\frac{x}{2} \right)\equiv \sin x\)
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\[2\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\left( \sin^2\frac{x}{2} \right)\equiv \sin x\]
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\[2\cos \frac{x}{2}\sin \frac{x}{2}\equiv \sin x\]
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\[\sin2A=2\sin A\cos A\]
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\[\sin x\equiv \sin x\]