Note: You need to have your calculator in radians (RAD) mode for this task (since the angles are given in radians - i.e., π is featured). a) A current waveform in a robotic arm may be described by... i_s=13 cos (2πft-(π/4))(A), where frequency, f=1Hz and t represents time. Make time (t) the subject of this formula and hence determine a point in time when the current waveform has a magnitude of +10A .

**Step 1. Write the equation for the current** We are given \[ i_s = 13\cos\left(2\pi ft - \frac{\pi}{4}\right) \quad [\text{A}] \] with \( f = 1 \) Hz. Therefore, the equation becomes \[ 13\cos\left(2\pi t - \frac{\pi}{4}\right) = 10 \] **Step 2. Isolate the cosine term** Divide both sides of the equation by 13: \[ \cos\left(2\pi t - \frac{\pi}{4}\right) = \frac{10}{13} \] **Step 3. Take the inverse cosine** The general solution for an equation \[ \cos(\theta) = \frac{10}{13} \] is given by \[ \theta = \cos^{-1}\left(\frac{10}{13}\right) + 2\pi k \quad \text{or} \quad \theta = -\cos^{-1}\left(\frac{10}{13}\right) + 2\pi k \] where \( k \) is any integer. Here, the angle \(\theta\) corresponds to \[ \theta = 2\pi t - \frac{\pi}{4} \] **Step 4. Solve for \( t \)** Substitute back \(\theta = 2\pi t - \frac{\pi}{4}\) into each solution. For the first solution: \[ 2\pi t - \frac{\pi}{4} = \cos^{-1}\left(\frac{10}{13}\right) + 2\pi k \] Add \(\frac{\pi}{4}\) to both sides: \[ 2\pi t = \cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4} + 2\pi k \] Divide by \(2\pi\): \[ t = \frac{\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} + k \] For the second solution: \[ 2\pi t - \frac{\pi}{4} = -\cos^{-1}\left(\frac{10}{13}\right) + 2\pi k \] Add \(\frac{\pi}{4}\) to both sides: \[ 2\pi t = -\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4} + 2\pi k \] Divide by \(2\pi\): \[ t = \frac{-\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} + k \] Thus, the general solutions for \( t \) are: \[ \boxed{t = \frac{\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} + k \quad \text{or} \quad t = \frac{-\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} + k, \quad k \in \mathbb{Z}} \] **Step 5. Determine a specific point in time** To find a specific time when the current is \(+10\) A, choose \( k = 0 \). For the first solution with \( k=0 \): \[ t = \frac{\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} \] For the second solution with \( k=0 \): \[ t = \frac{-\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}}{2\pi} \] Using approximate values: - Calculate \(\cos^{-1}\left(\frac{10}{13}\right)\). Since \(\frac{10}{13} \approx 0.7692\), we have \(\cos^{-1}(0.7692) \approx 0.698\) radians. - \(\frac{\pi}{4} \approx 0.7854\) radians. **First solution with \( k=0 \):** \[ t \approx \frac{0.698 + 0.7854}{2\pi} = \frac{1.4834}{6.2832} \approx 0.236\ \text{seconds} \] **Second solution with \( k=0 \):** \[ t \approx \frac{-0.698 + 0.7854}{2\pi} = \frac{0.0874}{6.2832} \approx 0.0139\ \text{seconds} \] Either time is a valid solution for when the current is \(+10\) A. One possible answer is \[ \boxed{t \approx 0.0139\ \text{seconds}} \] with the general expressions given above.