Further trigonometry

Trigonometry

Secant, cosecant and cotangent

$\sec\theta = \frac{1}{\cos\theta} \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$

Two more identities

Dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ and by $\sin^2\theta$ gives:

$2(1 + \tan^2 x) = 5\tan x \quad\Rightarrow\quad 2\tan^2 x - 5\tan x + 2 = 0 \quad\Rightarrow\quad (2\tan x - 1)(\tan x - 2) = 0$

$\tan x = \tfrac{1}{2}$ gives $x = 26.6^\circ$ or $206.6^\circ$. $\tan x = 2$ gives $x = 63.4^\circ$ or $243.4^\circ$.

The addition formulae

The signs flip in the cosine formula and the bottom of the tangent formula. These are on the formula list, but you must be able to use them quickly.

$\sin(45^\circ + 30^\circ) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = 2\cos x \quad\Rightarrow\quad \frac{\sqrt{3}}{2}\sin x = \frac{3}{2}\cos x \quad\Rightarrow\quad \tan x = \sqrt{3}$

So $x = 60^\circ$.

The double angle formulae

Putting $B = A$ in the addition formulae gives:

Choose the form of $\cos 2A$ that leaves only one ratio in the equation.

$1 - 2\sin^2 x + 3\sin x = 2 \quad\Rightarrow\quad 2\sin^2 x - 3\sin x + 1 = 0 \quad\Rightarrow\quad (2\sin x - 1)(\sin x - 1) = 0$

$\sin x = \tfrac{1}{2}$ gives $30^\circ$ or $150^\circ$. $\sin x = 1$ gives $90^\circ$.

$2\sin x\cos x - \cos x = 0$, so $\cos x(2\sin x - 1) = 0$. Then $\cos x = 0$ gives $90^\circ$ or $270^\circ$, and $\sin x = \tfrac{1}{2}$ gives $30^\circ$ or $150^\circ$.

$\frac{1 - (1 - 2\sin^2 x)}{2\sin x\cos x} = \frac{2\sin^2 x}{2\sin x\cos x} = \frac{\sin x}{\cos x} = \tan x$

Writing $a\sin\theta + b\cos\theta$ as a single wave

Any sum of a sine and a cosine of the same angle is one wave, written as $R\sin(\theta \pm \alpha)$ or $R\cos(\theta \pm \alpha)$, with $R > 0$ and $\alpha$ acute.

$R\sin(x + \alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha$. Matching gives $R\cos\alpha = 3$ and $R\sin\alpha = 4$.

So $R = 5$ and $\tan\alpha = \dfrac{4}{3}$, giving $\alpha = 53.13^\circ$:

$3\sin x + 4\cos x = 5\sin(x + 53.13^\circ)$

Using the single wave form

Solving equations. Solve $3\sin x + 4\cos x = 2$ for $0^\circ \leq x \leq 360^\circ$.

$\sin(x + 53.13^\circ) = 0.4$, in the interval $53.13^\circ \leq x + 53.13^\circ \leq 413.13^\circ$. The principal value is $23.58^\circ$, which is outside the interval. The values inside are $156.42^\circ$ and $383.58^\circ$, so $x = 103.3^\circ$ or $x = 330.4^\circ$.

Greatest and least values. $5\sin(x + 53.13^\circ)$ runs from $-5$ to $5$. The greatest value is reached when $x + 53.13^\circ = 90^\circ$, so at $x = 36.87^\circ$.

The denominator is $8 + 5\sin(x + 53.13^\circ)$, whose least value is $3$. So the greatest value of the fraction is $\dfrac{7}{3}$.