Let `f(x) = {{:(-2 sin x,"for",-pi le x le - (pi)/(2)),(a sin x + b,"for",-(pi)/(2) lt x lt (pi)/(2)),(cos x,"for",(pi)/(2) le x le pi):}`.
If `f` is continuous on `[-pi, pi)`, then find the values of `a` and `b`.

Answer

To find the values of \( a \) and \( b \) such that the piecewise function

\[ f(x) = \begin{cases} -2 \sin x & \text{for } -\pi \leq x \leq -\frac{\pi}{2} \\ a \sin x + b & \text{for } -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x & \text{for } \frac{\pi}{2} \leq x \leq \pi \end{cases} \]

is continuous on the interval \([- \pi, \pi)\), we need to ensure continuity at the points \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

### Step 1: Continuity at \( x = -\frac{\pi}{2} \)

To ensure continuity at \( x = -\frac{\pi}{2} \), we set the left-hand limit equal to the right-hand limit:

\[ \lim_{x \to -\frac{\pi}{2}^-} f(x) = \lim_{x \to -\frac{\pi}{2}^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to -\frac{\pi}{2}} (-2 \sin x) = -2 \sin\left(-\frac{\pi}{2}\right) = -2 \cdot (-1) = 2 \]

Calculating the right-hand limit:

\[ \lim_{x \to -\frac{\pi}{2}} (a \sin x + b) = a \sin\left(-\frac{\pi}{2}\right) + b = a \cdot (-1) + b = -a + b \]

Setting these equal gives us our first equation:

\[ 2 = -a + b \quad \text{(Equation 1)} \]

### Step 2: Continuity at \( x = \frac{\pi}{2} \)

Next, we ensure continuity at \( x = \frac{\pi}{2} \):

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to \frac{\pi}{2}} (a \sin x + b) = a \sin\left(\frac{\pi}{2}\right) + b = a \cdot 1 + b = a + b \]

Calculating the right-hand limit:

\[ \lim_{x \to \frac{\pi}{2}} (\cos x) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Setting these equal gives us our second equation:

\[ a + b = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

From Equation 2, we can express \( a \) in terms of \( b \):

\[ a = -b \]

Substituting this into Equation 1:

\[ 2 = -(-b) + b \]
\[ 2 = b + b \]
\[ 2 = 2b \]
\[ b = 1 \]

Now substituting \( b = 1 \) back into Equation 2 to find \( a \):

\[ a + 1 = 0 \implies a = -1 \]

### Conclusion

The values of \( a \) and \( b \) are:

\[ a = -1, \quad b = 1 \]