Question

1. for $\cos\theta = -\frac{1}{\sqrt{2}}$, determine $\theta$ if $0 \leq \theta \leq 2\pi$ . /5

1. prove the trig identity: $\frac{\sin2x}{1-\cos2x} = \cot x$ /5

Explanation:

Question 6:

Step1: Recall cosine values

We know that $\cos\theta = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$. The cosine function is negative in the second and third quadrants. The reference angle $\alpha$ where $\cos\alpha=\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$ (since $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$).

Step2: Find $\theta$ in second quadrant

In the second quadrant, $\theta = \pi - \alpha$. So $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Step3: Find $\theta$ in third quadrant

In the third quadrant, $\theta = \pi + \alpha$. So $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.

Step1: Use double - angle formulas

Recall the double - angle formulas: $\sin2x = 2\sin x\cos x$ and $\cos2x = 1 - 2\sin^{2}x$, so $1-\cos2x=2\sin^{2}x$.
Substitute these into the left - hand side (LHS) of the identity $\frac{\sin2x}{1 - \cos2x}$:
$\frac{\sin2x}{1 - \cos2x}=\frac{2\sin x\cos x}{2\sin^{2}x}$

Step2: Simplify the fraction

Cancel out the common factors. We can cancel out a factor of $2$ and a factor of $\sin x$ (assuming $\sin x
eq0$, which is valid because if $\sin x = 0$, then the left - hand side $\frac{\sin2x}{1 - \cos2x}=\frac{0}{1 - 1}$, which is undefined, and the right - hand side $\cot x=\frac{\cos x}{\sin x}$ is also undefined, so we can ignore the case $\sin x = 0$ for the domain of the identity).
After canceling, we get $\frac{\cos x}{\sin x}$.

Step3: Recall the definition of cotangent

We know that $\cot x=\frac{\cos x}{\sin x}$, so the left - hand side (LHS) is equal to the right - hand side (RHS).

Answer:

$\theta = \frac{3\pi}{4}, \frac{5\pi}{4}$

Question 7: