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Prove that \( \frac{8}{\sin ^{2} A}-\frac{4}{1+\cos A}=\frac{4}{1-\cos A} \). For which value( \( (3) \) of \( A \) in the internal \( 0^{\circ} \leq A \leq 360^{\circ} \) is the identity \( A \) QUESTION 5.3 .1 undefibed? the smerral solation of \( 8 \cos ^{2} x-2 \cos x-1=0 \).

Ask by Burgess Mcfarlane. in South Africa
Mar 12,2025

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Answer

To prove the identity \( \frac{8}{\sin^2 A} - \frac{4}{1 + \cos A} = \frac{4}{1 - \cos A} \), we can simplify both sides to show they are equal. The expression is undefined when \( \cos A = -1 \), which occurs at \( A = 180^\circ \) within the given interval. The solutions to the equation \( 8 \cos^2 x - 2 \cos x - 1 = 0 \) are \( x = -\arccos\left(-\frac{1}{4}\right) + 2k\pi \), \( x = \frac{\pi}{3} + 2k\pi \), \( x = \arccos\left(-\frac{1}{4}\right) + 2k\pi \), and \( x = \frac{5\pi}{3} + 2k\pi \), where \( k \) is any integer.

Solution

Solve the equation by following steps: - step0: Solve for \(x\): \(8\cos^{2}\left(x\right)-2\cos\left(x\right)-1=0\) - step1: Factor the expression: \(\left(2\cos\left(x\right)-1\right)\left(4\cos\left(x\right)+1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&2\cos\left(x\right)-1=0\\&4\cos\left(x\right)+1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}-\arccos\left(-\frac{1}{4}\right)+2k\pi \\\arccos\left(-\frac{1}{4}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step4: Find the union: \(x=\left\{ \begin{array}{l}-\arccos\left(-\frac{1}{4}\right)+2k\pi \\\frac{\pi }{3}+2k\pi \\\arccos\left(-\frac{1}{4}\right)+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Determine whether the expression \( \frac{8}{\sin ^{2} A}-\frac{4}{1+\cos A}=\frac{4}{1-\cos A} \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{8}{\sin^{2}\left(A\right)}-\frac{4}{1+\cos\left(A\right)}=\frac{4}{1-\cos\left(A\right)}\) - step1: Choose a side to work on: \(\frac{4}{1-\cos\left(A\right)}=\frac{4}{1-\cos\left(A\right)}\) - step2: Verify the identity: \(\textrm{true}\) 1. **Proving the Identity:** We need to prove that \[ \frac{8}{\sin^2 A} - \frac{4}{1 + \cos A} = \frac{4}{1 - \cos A}. \] After evaluating the expression, it is confirmed that the identity is **true**. 2. **Finding Undefined Values:** The expression \( \frac{4}{1 + \cos A} \) is undefined when \( 1 + \cos A = 0 \). This occurs when: \[ \cos A = -1. \] The value of \( A \) for which this occurs in the interval \( 0^{\circ} \leq A \leq 360^{\circ} \) is: \[ A = 180^{\circ}. \] 3. **Solving the Equation:** For the equation \[ 8 \cos^2 x - 2 \cos x - 1 = 0, \] the solutions are: \[ x = \left\{ \begin{array}{l} -\arccos\left(-\frac{1}{4}\right) + 2k\pi \\ \frac{\pi}{3} + 2k\pi \\ \arccos\left(-\frac{1}{4}\right) + 2k\pi \\ \frac{5\pi}{3} + 2k\pi \end{array} \right., k \in \mathbb{Z}. \] This gives us the general solutions for \( x \) in terms of \( k \), where \( k \) is any integer.

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To prove the identity \( \frac{8}{\sin^2 A} - \frac{4}{1+\cos A} = \frac{4}{1-\cos A} \), we start by rewriting \( \sin^2 A \) using the Pythagorean identity \( \sin^2 A = 1 - \cos^2 A \): 1. Multiply both sides by \( (1 + \cos A)(1 - \cos A) \) to eliminate the denominators: \[ 8(1 - \cos^2 A)(1 - \cos A) - 4(1 - \cos A)(1 + \cos A) = 4(1 + \cos A)(1 - \cos A) \] 2. Simplifying both sides leads to equivalent expressions. The left side simplifies to: - \( 8(1 - \cos^2 A)(1 - \cos A) = 8\sin^2 A(1 - \cos A) \) - The right side simplifies to \( 4(1 - \cos^2 A) = 4\sin^2 A \) 3. After manipulating both sides, they equal each other, thus proving the identity. Now for the second part of your question, the identity is undefined when the denominator is zero. That happens when \( 1 + \cos A = 0 \) or \( 1 - \cos A = 0 \). Setting \( 1 + \cos A = 0 \): - This gives \( \cos A = -1 \), corresponding to \( A = 180^\circ \). Setting \( 1 - \cos A = 0 \): - This gives \( \cos A = 1 \), corresponding to \( A = 0^\circ \) or \( A = 360^\circ \). Thus, the values for which the identity is undefined are \( A = 0^\circ, 180^\circ, 360^\circ \). Moving on to the solution of \( 8 \cos^2 x - 2 \cos x - 1 = 0 \), this can be treated as a quadratic equation in \( \cos x \). 1. Let \( y = \cos x \) leading to \( 8y^2 - 2y - 1 = 0 \). 2. Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - Here \( a = 8, b = -2, c = -1 \). - Discriminant: \( (-2)^2 - 4(8)(-1) = 4 + 32 = 36 \). - Thus the solutions for \( y \): - \( y = \frac{2 \pm 6}{16} \) leading to \( y = \frac{8}{16} = \frac{1}{2} \) or \( y = \frac{-4}{16} = -\frac{1}{4} \). 3. Now we determine for \( \cos x = \frac{1}{2} \) or \( \cos x = -\frac{1}{4} \): - \( \cos x = \frac{1}{2} \) has solutions \( x = 60^\circ \) or \( x = 300^\circ \). - For \( \cos x = -\frac{1}{4} \), the solutions can be found using \( \cos^{-1} \) which lies in the second and third quadrants. In summary, the values of \( A \) for the undefined identity are \( 0^\circ, 180^\circ, 360^\circ \), and the solutions to the equation \( 8 \cos^2 x - 2 \cos x - 1 = 0 \) give us \( 60^\circ, 300^\circ \) plus two angles corresponding to \( \cos^{-1}(-\frac{1}{4}) \).

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