Express \(\frac{1}{2} \cos t+\sin t\) in the form \(A \sin (\omega t-\alpha), \alpha \geq 0\)

Recall the sine addition formula: \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\). We want to rewrite \(\frac{1}{2} \cos t + \sin t\) as a sine with a phase shift: \(A \sin (\omega t - \alpha)\). Let's compare the given expression with the sine addition formula: \(\frac{1}{2} \cos t + \sin t = A\sin(\omega t)\cos(-\alpha) + A\cos(\omega t)\sin(-\alpha)\). It becomes clear that we should take \(\omega =1\), so that coefficients of \(\cos t\) and \(\sin t\) match on both sides of the equation above.

Now, our task is to find \(A\) and \(\alpha\): \(\frac{1}{2} \cos t + \sin t = A\sin t\cos(-\alpha) + A\cos t\sin(-\alpha)\). For the coefficients to match, we need: \(\frac{1}{2} = A\sin(-\alpha) \quad(1)\) \(1 = A\cos(-\alpha) \quad(2)\)

Square both equations (1) and (2) and add them: \((\frac{1}{2})^2 + 1^2 = (A\sin(-\alpha))^2 + (A\cos(-\alpha))^2\) \(\frac{1}{4} + 1 = A^2 (\sin^2(-\alpha) + \cos^2(-\alpha))\) \(\frac{5}{4} = A^2\). Taking the square root of both sides, we obtain: \(A = \frac{\sqrt{5}}{2}\). Now, let's solve for \(\alpha\). Divide equation (1) by (2): \(\frac{1/2}{1} = \frac{\sin(-\alpha)}{\cos(-\alpha)}\) \(\tan(-\alpha) = \frac{1}{2}\) \(\alpha = -\arctan\left(\frac{1}{2}\right)\). For \(\alpha\) to satisfy \(\alpha \geq 0\), we can add \(\pi\) to the obtained value: \(\alpha = \pi - \arctan\left(\frac{1}{2}\right)\). Finally, we found all the necessary values and can express our original function in the desired form:

The function \(\frac{1}{2} \cos t + \sin t\) can be expressed in the form \(A \sin (\omega t - \alpha)\) with the found values for \(A\), \(\omega\), and \(\alpha\): \(\frac{1}{2} \cos t + \sin t = \frac{\sqrt{5}}{2} \sin (t - (\pi - \arctan\left(\frac{1}{2}\right)))\).