Trigonometric Functions, Identities and Equations

Degrees and radians

A full turn is $360°$ or $2\pi$ radians, so a half turn is $180° = \pi$ radians.

So $135° = 135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}$. Cancel the fraction before you finish — $\tfrac{135}{180}$ reduces to $\tfrac34$.

Worth memorising: $30° = \tfrac{\pi}{6}$, $45° = \tfrac{\pi}{4}$, $60° = \tfrac{\pi}{3}$, $90° = \tfrac{\pi}{2}$.

The three graphs

Sine and cosine are the same wave, shifted by $90°$. Both oscillate between $-1$ and $1$, which is why an equation like $\sin\theta = 2$ has no solution at all.

Tangent is different: it repeats every $180°$, and it shoots off to infinity at $90°$ and $270°$ where $\cos\theta = 0$ and the division is undefined.

Where each ratio is positive

Divide the turn into four quadrants and remember which ratios are positive in each:

• $0°$–$90°$: all three are positive
• $90°$–$180°$: only sine
• $180°$–$270°$: only tangent
• $270°$–$360°$: only cosine

Reading the initials anticlockwise from the fourth quadrant spells CAST, which is how most people remember it.

This is what tells you where the second solution lives. If $\sin\theta$ is positive, $\theta$ is in the first or second quadrant; if $\cos\theta$ is negative, $\theta$ is in the second or third.

The identities

The first is Pythagoras on a triangle with hypotenuse 1. Rearranged, it is the workhorse of this whole section:

$$\sin^2\theta = 1 - \cos^2\theta \qquad\text{and}\qquad \cos^2\theta = 1 - \sin^2\theta$$

Use it whenever an equation contains both $\sin^2\theta$ and $\cos\theta$ (or both $\cos^2\theta$ and $\sin\theta$) — substituting turns the equation into a quadratic in one ratio.

Proving an identity

Start with the messier side and work towards the other. Do not move terms across the $=$ as if it were an equation — you are showing the two sides are already equal, not solving for anything.

The three moves that solve almost every one:

1. Replace $1 - \cos^2\theta$ with $\sin^2\theta$, or the other way round.
2. Turn $\tan\theta$ into $\dfrac{\sin\theta}{\cos\theta}$.
3. Put everything over a common denominator and simplify.

Solving trigonometric equations

1. Rearrange until you have a single ratio equal to a number.
2. Find the principal value — the first angle that works.
3. Use the quadrant rule to find every other solution inside the given interval.
4. Check each one is actually in the interval before writing it down.

For $2\sin\theta = 1$ over $0° \leq \theta \leq 360°$: $\sin\theta = \tfrac12$, so the principal value is $30°$. Sine is also positive in the second quadrant, giving $180° - 30° = 150°$.

When the equation is quadratic in one ratio, substitute a letter for it, solve the quadratic, then solve each resulting equation separately. Remember to discard any value outside $-1 \leq k \leq 1$ for sine or cosine — it cannot come from a real angle.