Test of Mathematics for University Admission (TMUA) Trigonometry Syllabus

What are the period, amplitude and range of $y = \cos x$?

Period $2\pi$ (or $360^{\circ}$), amplitude $1$, range $-1 \leq y \leq 1$. It has a maximum of $1$ at $x = 0$ and equals $0$ at $x = \tfrac{\pi}{2} + n\pi$.

Describe the key features of the graph of $y = \tan x$.

Period $\pi$ (or $180^{\circ}$), range all real numbers, passes through the origin, and has vertical asymptotes at $x = \tfrac{\pi}{2} + n\pi$ where $\cos x = 0$.

How is the graph of $y = \cos x$ related to $y = \sin x$ by a translation?

$\cos x = \sin\left(x + \tfrac{\pi}{2}\right)$, so the cosine graph is the sine graph translated $\tfrac{\pi}{2}$ (i.e. $90^{\circ}$) to the left.

State whether $\sin$, $\cos$ and $\tan$ are odd or even functions, with the defining identity.

$\sin$ is odd: $\sin(-x) = -\sin x$. $\cos$ is even: $\cos(-x) = \cos x$. $\tan$ is odd: $\tan(-x) = -\tan x$.

How does the graph of $y = a\sin x$ differ from $y = \sin x$ (for $a > 0$)?

It is a vertical stretch by scale factor $a$, giving amplitude $a$ and range $-a \leq y \leq a$; the period is unchanged.

How does the graph of $y = \sin(bx)$ differ from $y = \sin x$ (for $b > 0$)?

It is a horizontal stretch by scale factor $\tfrac{1}{b}$, giving period $\dfrac{2\pi}{b}$ (or $\dfrac{360^{\circ}}{b}$); the amplitude is unchanged.

Describe the transformation taking $y = f(x)$ to $y = f(x - c)$ and to $y = f(x) + d$.

$y = f(x - c)$ is a translation by $c$ in the positive $x$-direction (right). $y = f(x) + d$ is a translation by $d$ in the positive $y$-direction (up).

What transformations map $y = \cos x$ onto $y = \cos(-x)$ and onto $y = -\cos x$?

$y = \cos(-x) = \cos x$ (reflection in the $y$-axis leaves it unchanged, as cosine is even). $y = -\cos x$ is a reflection in the $x$-axis.

For $y = a\sin(bx + c) + d$, identify the amplitude, period and vertical shift.

Amplitude $= |a|$, period $= \dfrac{2\pi}{|b|}$, and vertical shift (midline) $= d$ so the range is $d - |a| \leq y \leq d + |a|$.

State the Pythagorean identity relating $\sin$ and $\cos$.

$\sin^{2}\theta + \cos^{2}\theta = 1$.

Rearrange $\sin^{2}\theta + \cos^{2}\theta = 1$ to express $\sin^{2}\theta$ and $\cos^{2}\theta$ individually.

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

What is the identity for $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$, and where is it valid?

$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, valid for $\cos\theta \neq 0$, i.e. $\theta \neq \tfrac{\pi}{2} + n\pi$.

In which quadrants are $\sin$, $\cos$ and $\tan$ positive (the CAST rule)?

All positive in quadrant 1; only $\sin$ positive in quadrant 2; only $\tan$ positive in quadrant 3; only $\cos$ positive in quadrant 4 (remembered as CAST going anticlockwise from Q4).

Give the symmetry identities relating $\sin(180^{\circ} - \theta)$ and $\cos(180^{\circ} - \theta)$ to $\theta$.

$\sin(180^{\circ} - \theta) = \sin\theta$ and $\cos(180^{\circ} - \theta) = -\cos\theta$.

State the complementary-angle identities relating sine and cosine.

$\sin(90^{\circ} - \theta) = \cos\theta$ and $\cos(90^{\circ} - \theta) = \sin\theta$ (equivalently $\sin\left(\tfrac{\pi}{2} - \theta\right) = \cos\theta$).

When solving $\sin\theta = k$ for $0^{\circ} \leq \theta < 360^{\circ}$, how do you find the second solution from the principal value $\alpha$?

The two solutions are $\theta = \alpha$ and $\theta = 180^{\circ} - \alpha$ (then add multiples of $360^{\circ}$ to find others in range).

When solving $\cos\theta = k$ for $0^{\circ} \leq \theta < 360^{\circ}$, how do you find the second solution from the principal value $\alpha$?

The solutions are $\theta = \alpha$ and $\theta = 360^{\circ} - \alpha$ (i.e. $\pm\alpha$ plus multiples of $360^{\circ}$).

When solving $\tan\theta = k$, how are successive solutions related?

Solutions repeat every $180^{\circ}$ (or $\pi$): if $\theta = \alpha$ is one solution, then $\theta = \alpha + 180^{\circ}n$ for integer $n$.

Outline the process for solving an equation such as $\sin(2x) = k$ over a given interval for $x$.

Multiply the interval for $x$ by the coefficient to get the interval for the argument $u = 2x$, solve $\sin u = k$ for all $u$ in that wider range, then divide each solution by $2$ to recover $x$.

How can $2\cos^{2}\theta + \cos\theta - 1 = 0$ be solved using an identity or substitution?

It is already quadratic in $\cos\theta$: let $c = \cos\theta$, solve $2c^{2} + c - 1 = 0$ to get $c = \tfrac{1}{2}$ or $c = -1$, then solve $\cos\theta = \tfrac{1}{2}$ and $\cos\theta = -1$.

How do you solve an equation mixing $\sin^{2}\theta$ and $\cos\theta$, such as $2\sin^{2}\theta = 1 + \cos\theta$?

Use $\sin^{2}\theta = 1 - \cos^{2}\theta$ to write everything in $\cos\theta$, giving a quadratic in $\cos\theta$, then solve.

State the compound-angle formulae for $\sin(A \pm B)$.

$\sin(A + B) = \sin A\cos B + \cos A\sin B$ and $\sin(A - B) = \sin A\cos B - \cos A\sin B$.

State the compound-angle formulae for $\cos(A \pm B)$.

$\cos(A + B) = \cos A\cos B - \sin A\sin B$ and $\cos(A - B) = \cos A\cos B + \sin A\sin B$. (Note the sign 'swap' relative to the sine formula.)

State the compound-angle formula for $\tan(A \pm B)$.

$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$.

State the double-angle formula for $\sin 2A$.

$\sin 2A = 2\sin A\cos A$.

State the three equivalent forms of the double-angle formula for $\cos 2A$.

$\cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A$.

State the double-angle formula for $\tan 2A$.

$\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$.

How can $\cos 2A = 2\cos^{2}A - 1$ be rearranged to give $\cos^{2}A$ (the power-reduction form)?

$\cos^{2}A = \dfrac{1 + \cos 2A}{2}$, and similarly $\sin^{2}A = \dfrac{1 - \cos 2A}{2}$.

Using a double-angle identity, what does $\sin\theta\cos\theta$ simplify to?

$\sin\theta\cos\theta = \dfrac{1}{2}\sin 2\theta$ (from $\sin 2\theta = 2\sin\theta\cos\theta$).

Why must you check candidate solutions of a trigonometric equation against the stated interval and any domain restrictions?

Squaring, dividing by a trig term, or using identities can introduce extraneous solutions or lose solutions (e.g. where $\cos\theta = 0$ for $\tan$); each candidate must lie in the given interval and satisfy the original equation.