Trigonometry — practice questions

The equation contains both $\tan^2\theta$ and $\sec\theta$. We need to express it in terms of a single trigonometric function. We use the identity $1 + \tan^2\theta \equiv \sec^2\theta$, which means $\tan^2\theta \equiv \sec^2\theta - 1$.

Substitute this into the equation: $2(\sec^2\theta - 1) = 3 + \sec\theta$ $2\sec^2\theta - 2 = 3 + \sec\theta$

This is a quadratic equation in $\sec\theta$. Let's rearrange it into standard form: $2\sec^2\theta - \sec\theta - 5 = 0$

This doesn't factorise easily, so we use the quadratic formula for $y = \sec\theta$, where $a=2, b=-1, c=-5$: $\sec\theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-5)}}{2(2)}$ $\sec\theta = \frac{1 \pm \sqrt{1 + 40}}{4} = \frac{1 \pm \sqrt{41}}{4}$

This gives two possible values for $\sec\theta$: $\sec\theta \approx 1.8508$ or $\sec\theta \approx -1.3508$

Now we convert back to $\cos\theta$ using $\cos\theta = 1/\sec\theta$: $\cos\theta \approx 1/1.8508 \approx 0.5403$ or $\cos\theta \approx 1/(-1.3508) \approx -0.7403$

For $\cos\theta \approx 0.5403$: Principal value $\theta = \arccos(0.5403) \approx 57.3^\circ$. Cosine is positive in the 1st and 4th quadrants. The second solution is $360^\circ - 57.3^\circ = 302.7^\circ$.

For $\cos\theta \approx -0.7403$: Principal value $\theta = \arccos(-0.7403) \approx 137.7^\circ$. Cosine is negative in the 2nd and 3rd quadrants. The second solution is $360^\circ - 137.7^\circ = 222.3^\circ$.

So, the solutions in the range $0^\circ \le \theta \le 360^\circ$ are approximately $57.3^\circ, 137.7^\circ, 222.3^\circ, 302.7^\circ$ (to 1 d.p.).

a) We want to match $5\sin x - 12\cos x$ with $R\sin(x - \alpha) \equiv R(\sin x \cos \alpha - \cos x \sin \alpha)$. By comparing coefficients of $\sin x$ and $\cos x$: $R\cos\alpha = 5$ (1) $R\sin\alpha = 12$ (2)

To find R, we square and add the two equations: $(R\cos\alpha)^2 + (R\sin\alpha)^2 = 5^2 + 12^2$ $R^2(\cos^2\alpha + \sin^2\alpha) = 25 + 144$ $R^2(1) = 169$ $R = 13$ (since $R>0$)

To find $\alpha$, we divide equation (2) by (1): $\frac{R\sin\alpha}{R\cos\alpha} = \frac{12}{5}$ $\tan\alpha = 2.4$ $\alpha = \arctan(2.4) \approx 67.38^\circ$

So, $5\sin x - 12\cos x \equiv 13\sin(x - 67.38^\circ)$.

b) The expression is $13\sin(x - 67.38^\circ)$. The maximum value of the sine function is 1. Therefore, the maximum value of the expression is $13 \times 1 = 13$.

This occurs when $\sin(x - 67.38^\circ) = 1$. Let $Y = x - 67.38^\circ$. We need $\sin Y = 1$. The principal value is $Y = 90^\circ$. So, $x - 67.38^\circ = 90^\circ$. $x = 90^\circ + 67.38^\circ = 157.38^\circ$. This is the smallest positive value of $x$. The next would be at $90^\circ + 360^\circ$, which gives a larger $x$.

Maximum value: 13. Smallest positive $x$: $157.4^\circ$ (to 1 d.p.).