Trigonometry — practice questions

The condition $\pi < \theta < \frac{3\pi}{2}$ means the angle $\theta$ is in the third quadrant. In this quadrant, both sine and cosine are negative, while tangent is positive.

Step 1: Use the Pythagorean identity to find $\sin \theta$. We know that $\sin^2 \theta + \cos^2 \theta = 1$. Substituting the given value of $\cos \theta$: $\sin^2 \theta + \left(-\frac{4}{5}\right)^2 = 1$ $\sin^2 \theta + \frac{16}{25} = 1$ $\sin^2 \theta = 1 - \frac{16}{25} = \frac{9}{25}$ $\sin \theta = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$

Step 2: Determine the sign of $\sin \theta$. Since $\theta$ is in the third quadrant, $\sin \theta$ must be negative. [M1 for using quadrant] Therefore, $\sin \theta = -\frac{3}{5}$. [A1]

Step 3: Use the identity for $\tan \theta$. We know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$. $\tan \theta = \frac{-3/5}{-4/5}$ $\tan \theta = \frac{3}{4}$. [A1]

Final Answer: $\sin \theta = -\frac{3}{5}$ and $\tan \theta = \frac{3}{4}$.

Step 1: Rearrange the equation to isolate the trigonometric function. $2\sin^2 x - 1 = 0$ $2\sin^2 x = 1$ $\sin^2 x = \frac{1}{2}$

Step 2: Take the square root of both sides. Remember to include both the positive and negative roots. $\sin x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$ [M1 for isolating sin x correctly]

Step 3: Solve for the positive case, $\sin x = \frac{1}{\sqrt{2}}$. First, find the principal value (the acute angle). This is a standard exact value. $x = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$. Since sine is positive in the first and second quadrants, the second solution is: $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. [A1 for both solutions]

Step 4: Solve for the negative case, $\sin x = -\frac{1}{\sqrt{2}}$. First, find the related acute angle, which is still $\frac{\pi}{4}$. Sine is negative in the third and fourth quadrants. The third quadrant solution is $x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$. The fourth quadrant solution is $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$. [A1 for both solutions]

Step 5: Combine all solutions. The solutions in the range $0 \le x \le 2\pi$ are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.