Let's go through each part of the problem and solve it:

(i) Given $\sin \theta = -\frac{3}{5}$, with $180^\circ \leq \theta \leq 270^\circ$, find $\cos \theta$.

• First, recognize that $\theta$ is in the third quadrant (since $180^\circ \leq \theta \leq 270^\circ$).

• In the third quadrant, $\sin \theta$ is negative, and $\cos \theta$ is also negative.

• Since $\sin^2 \theta + \cos^2 \theta = 1$, we use the identity: $\sin^2 \theta + \cos^2 \theta = 1$ Substituting $\sin \theta = -\frac{3}{5}$: $\left( -\frac{3}{5} \right)^2 + \cos^2 \theta = 1$ $\frac{9}{25} + \cos^2 \theta = 1$ $\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$ $\cos \theta = \pm \frac{4}{5}$ Since $\cos \theta$ is negative in the third quadrant, we have: $\cos \theta = -\frac{4}{5}$

(ii) Determine the quadrant where $\sin \theta < 0$ and $\tan \theta > 0$.

• If $\sin \theta < 0$, then $\theta$ is either in the third or fourth quadrant.
• If $\tan \theta > 0$, then $\theta$ is in a quadrant where both sine and cosine have the same sign. This happens in the third quadrant, where both sine and cosine are negative, and their ratio (tangent) is positive.

Thus, $\theta$ lies in the third quadrant.

(iii) If $f(x) = x^2 + 2x - 1$, find $f(0)$.

• To find $f(0)$, substitute $x = 0$ into the function: $f(0) = (0)^2 + 2(0) - 1 = -1$ Thus, $f(0) = -1$.

Would you like further clarification or details on any of these parts?

Here are 5 related questions for practice:

1. Find $\cos \theta$ if $\sin \theta = \frac{5}{13}$ and $\theta$ is in the second quadrant.
2. In which quadrant is $\theta$ if $\sin \theta > 0$ and $\cos \theta < 0$?
3. If $\tan \theta = -1$, in which quadrant is $\theta$?
4. For $f(x) = 3x^2 - 5x + 4$, find $f(2)$.
5. Solve for $x$ if $f(x) = 0$ for the function $f(x) = x^2 - 6x + 9$.

Tip: When using trigonometric identities like $\sin^2 \theta + \cos^2 \theta = 1$, always consider the quadrant to determine the correct sign of the values.