5.1 Prove the identity: \( \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} \) 5.2 Prove that: \( \cos \left(30^{\circ}+\theta\right)-\cos \left(30^{\circ}-\theta\right)=-\sin \theta \) 5.3 Consider the Geometric series: \( \cos \alpha+\sin 2 \alpha+4 \sin ^{2} \alpha \cdot \cos \alpha+\ldots \), where \( \alpha \) is an acute angle. Determine the value of \( r \), the constant ratio of the series.

### 5.1 Prove the identity We start with the left-hand side: \[ \frac{1-\sin 2A}{\sin A-\cos A} \] Recall that the double-angle formula for sine is: \[ \sin 2A = 2\sin A\cos A. \] Substitute this into the numerator: \[ 1-\sin 2A = 1-2\sin A\cos A. \] Notice that using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), we can write: \[ 1 - 2\sin A\cos A = \sin^2 A + \cos^2 A - 2\sin A\cos A. \] The right-hand side can be recognized as a perfect square: \[ \sin^2 A + \cos^2 A - 2\sin A\cos A = (\sin A-\cos A)^2. \] Thus, the original expression becomes: \[ \frac{(\sin A-\cos A)^2}{\sin A-\cos A}. \] Provided that \(\sin A-\cos A \neq 0\), we cancel one factor: \[ \frac{(\sin A-\cos A)^2}{\sin A-\cos A} = \sin A-\cos A. \] This completes the proof: \[ \frac{1-\sin 2A}{\sin A-\cos A} = \sin A-\cos A. \] --- ### 5.2 Prove that \[ \cos\left(30^{\circ}+\theta\right)-\cos\left(30^{\circ}-\theta\right)=-\sin\theta. \] We use the cosine subtraction formula: \[ \cos(x+y) - \cos(x-y) = -2\sin x \sin y. \] Let \( x = 30^\circ \) and \( y = \theta \). Then, \[ \cos\left(30^{\circ}+\theta\right)-\cos\left(30^{\circ}-\theta\right) = -2\sin 30^{\circ}\sin\theta. \] Since \(\sin 30^{\circ} = \frac{1}{2}\), we have: \[ -2\sin 30^{\circ}\sin\theta = -2\left(\frac{1}{2}\right)\sin\theta = -\sin\theta. \] Thus, we obtain: \[ \cos\left(30^{\circ}+\theta\right)-\cos\left(30^{\circ}-\theta\right) = -\sin\theta. \] --- ### 5.3 Determine the constant ratio \(r\) of the geometric series Consider the series: \[ \cos\alpha + \sin 2\alpha + 4\sin^2\alpha\cos\alpha + \ldots \] A geometric series has the form: \[ a,\quad ar,\quad ar^2,\quad \ldots \] where \(a\) is the first term and \(r\) is the common ratio. The first term is: \[ a = \cos\alpha. \] The second term is: \[ ar = \sin 2\alpha. \] Thus, the common ratio \(r\) is: \[ r = \frac{\sin 2\alpha}{\cos\alpha}. \] Recall the double-angle identity: \[ \sin 2\alpha = 2\sin\alpha\cos\alpha. \] Substitute this into the expression for \(r\): \[ r = \frac{2\sin\alpha\cos\alpha}{\cos\alpha}. \] Cancel \(\cos\alpha\) (assuming \(\cos\alpha \neq 0\)): \[ r = 2\sin\alpha. \] Thus, the constant ratio of the series is: \[ r = 2\sin\alpha. \]