Trigonometry — practice questions

The equation contains two different trigonometric functions, $\cot\theta$ and $\csc\theta$. We need to express one in terms of the other using an identity.

1. Use a Pythagorean Identity: Recall the identity $1 + \cot^2\theta \equiv \csc^2\theta$. We can rearrange this to $\cot^2\theta \equiv \csc^2\theta - 1$.
2. Substitute: Substitute this into the original equation: $2(\csc^2\theta - 1) - 5\csc\theta = 1$ $2\csc^2\theta - 2 - 5\csc\theta = 1$
3. Form a Quadratic: Rearrange into a quadratic equation in terms of $\csc\theta$. Let $y = \csc\theta$. $2\csc^2\theta - 5\csc\theta - 3 = 0$ $2y^2 - 5y - 3 = 0$
4. Solve the Quadratic: Factorise the quadratic equation. $(2y+1)(y-3) = 0$ So, $y = -\frac{1}{2}$ or $y = 3$.
5. Solve for $\theta$: Substitute back $\csc\theta = y$ and then $\sin\theta = 1/y$. Case 1: $\csc\theta = -\frac{1}{2} \implies \sin\theta = -2$. This has no solutions, as $-1 \le \sin\theta \le 1$. Case 2: $\csc\theta = 3 \implies \sin\theta = \frac{1}{3}$.
6. Find all solutions in the range: Principal value: $\theta = \arcsin(\frac{1}{3}) \approx 19.47^\circ$. The sine function is positive in the first and second quadrants. First quadrant solution: $\theta_1 = 19.5^\circ$ (to 1 d.p.) Second quadrant solution: $\theta_2 = 180^\circ - 19.47^\circ \approx 160.5^\circ$ (to 1 d.p.)

Final solutions are $\theta = 19.5^\circ$ and $\theta = 160.5^\circ$.

(i) Express in Harmonic Form

1. Set up the identity: We want $5\cos x - 2\sin x \equiv R\cos(x+\alpha)$.
2. Expand the right side: Using the compound angle formula, $R\cos(x+\alpha) = R(\cos x \cos\alpha - \sin x \sin\alpha) = (R\cos\alpha)\cos x - (R\sin\alpha)\sin x$.
3. Compare coefficients: Coefficient of $\cos x$: $5 = R\cos\alpha$ Coefficient of $\sin x$: $2 = R\sin\alpha$ (Note the signs match up: $-2\sin x$ and $-(R\sin\alpha)\sin x$)
4. Find R: Square and add the two equations: $R^2\cos^2\alpha + R^2\sin^2\alpha = 5^2 + 2^2$ $R^2(\cos^2\alpha + \sin^2\alpha) = 25 + 4 = 29$ $R^2 = 29 \implies R = \sqrt{29}$ (since $R>0$).
5. Find $\alpha$: Divide the equations: $\frac{R\sin\alpha}{R\cos\alpha} = \tan\alpha = \frac{2}{5}$ $\alpha = \arctan(0.4) \approx 21.8014^\circ$. To 1 d.p., $\alpha = 21.8^\circ$.

So, $5\cos x - 2\sin x \equiv \sqrt{29}\cos(x+21.8^\circ)$.

(ii) Solve the Equation

1. Substitute the harmonic form: The equation $5\cos x - 2\sin x = 4$ becomes: $\sqrt{29}\cos(x+21.8014^\circ) = 4$
2. Isolate the cosine term: $\cos(x+21.8014^\circ) = \frac{4}{\sqrt{29}}$
3. Adjust the interval: The interval for $x$ is $0^\circ \le x \le 360^\circ$. So the interval for our new angle, let's call it $u = x+21.8014^\circ$, is $21.8014^\circ \le u \le 381.8014^\circ$.
4. Find the principal value for u: $u = \arccos\left(\frac{4}{\sqrt{29}}\right) \approx 42.031^\circ$. This value is within our adjusted interval.
5. Find other solutions for u: Cosine is positive in the first and fourth quadrants. The other solution is found by $360^\circ - \text{principal value}$. $u_1 = 42.031^\circ$ $u_2 = 360^\circ - 42.031^\circ = 317.969^\circ$. Both $u_1$ and $u_2$ are within the interval $[21.8^\circ, 381.8^\circ]$.
6. Solve for x: Convert back using $x = u - 21.8014^\circ$. $x_1 = 42.031^\circ - 21.8014^\circ = 20.2296^\circ \approx 20.2^\circ$. $x_2 = 317.969^\circ - 21.8014^\circ = 296.1676^\circ \approx 296.2^\circ$.

Final solutions are $x = 20.2^\circ$ and $x = 296.2^\circ$.