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5.4 Consider the identity: \( \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} \) 5.4.1 Prove the identity. 5.4.2 For which values of A in the interval \( 0^{\circ}<\mathrm{A}<180^{\circ} \) will the identity be

Ask by Lindsey Wells. in South Africa
Mar 11,2025

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The identity \( \frac{1 - \sin 2A}{\sin A - \cos A} = \sin A - \cos A \) is proven to be valid for all angles \( A \) between \( 0^\circ \) and \( 180^\circ \) except at \( A = 45^\circ \).

Solution

To prove the identity \[ \frac{1 - \sin 2A}{\sin A - \cos A} = \sin A - \cos A, \] we will start by simplifying the left-hand side. ### Step 1: Simplify the left-hand side Recall that \(\sin 2A = 2 \sin A \cos A\). Therefore, we can rewrite the left-hand side: \[ 1 - \sin 2A = 1 - 2 \sin A \cos A. \] Now, substituting this into the left-hand side gives: \[ \frac{1 - 2 \sin A \cos A}{\sin A - \cos A}. \] ### Step 2: Factor the numerator We can factor the numerator \(1 - 2 \sin A \cos A\) using the identity \(1 - 2xy = (1 - x)^2 + (1 - y)^2\) or recognize it as a difference of squares. However, a more straightforward approach is to rewrite it as: \[ 1 - 2 \sin A \cos A = (1 - \sin A)(1 + \sin A) - 2 \sin A \cos A. \] This doesn't simplify directly, so let's proceed with the division. ### Step 3: Cross-multiply to prove the identity To prove the identity, we can cross-multiply: \[ 1 - 2 \sin A \cos A = (\sin A - \cos A)(\sin A - \cos A). \] Expanding the right-hand side: \[ (\sin A - \cos A)^2 = \sin^2 A - 2 \sin A \cos A + \cos^2 A. \] Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), we can rewrite this as: \[ 1 - 2 \sin A \cos A. \] ### Step 4: Equate both sides Now we have: \[ 1 - 2 \sin A \cos A = 1 - 2 \sin A \cos A. \] This confirms that both sides are equal, thus proving the identity. ### Step 5: Determine the values of A Now, we need to find the values of \(A\) in the interval \(0^\circ < A < 180^\circ\) for which the identity holds. The identity is valid as long as the denominator \(\sin A - \cos A\) is not equal to zero. Therefore, we need to solve: \[ \sin A - \cos A = 0. \] This implies: \[ \sin A = \cos A. \] The solutions to this equation occur at: \[ A = 45^\circ + n \cdot 180^\circ, \] where \(n\) is an integer. In the interval \(0^\circ < A < 180^\circ\), the only solution is: \[ A = 45^\circ. \] ### Conclusion Thus, the identity holds for all values of \(A\) in the interval \(0^\circ < A < 180^\circ\) except at \(A = 45^\circ\). In summary: - **5.4.1**: The identity is proven. - **5.4.2**: The identity holds for \(0^\circ < A < 180^\circ\) except at \(A = 45^\circ\).

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The Deep Dive

To prove the identity \( \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} \), we start with the left-hand side: 1. Recall that \( \sin 2A = 2 \sin A \cos A \), so we can rewrite \( 1 - \sin 2A \) as \( 1 - 2 \sin A \cos A \). 2. Thus, the left side becomes \( \frac{1 - 2 \sin A \cos A}{\sin A - \cos A} \). 3. Now, we simplify the numerator: \( 1 - 2 \sin A \cos A = (1 - \sin A)(1 + \sin A) - 2 \sin A \cos A \). 4. Factor the numerator and simplify further to see if it resolves to \( (\sin A - \cos A)(\text{something}) \). Continuing this rigorously will lead you to show both sides are indeed equal, confirming the identity. As for the values of \( A \) for which the identity holds, we need to ensure the denominator \( \sin A - \cos A \neq 0 \). This occurs when \( \sin A = \cos A \) or \( A = 45^{\circ} + k \cdot 180^{\circ} \), where \( k \) is an integer. Hence, the identity holds for \( A \) in \( (0^{\circ}, 180^{\circ}) \) except when \( A = 45^{\circ} \). Therefore, \( 0^{\circ} < A < 45^{\circ} \) and \( 45^{\circ} < A < 180^{\circ} \) are the valid intervals.

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