GCSE Exact trigonometric values Practice Questions

Derive the 30 and 60 degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Write the ratio for sin 30 degrees using the sides of that triangle.

For the $30^\circ$ angle in that triangle, opposite over hypotenuse gives $\frac{1}{2}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{1}{2}$, in simplest form with a rational denominator.

Derive the 30 and 60 degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Evaluate the option that works.

$\cos 60^\circ$ comes out as exactly $\frac{1}{2}$, which is the value asked for.

Select the correct option.

$\cos 60^\circ$ is the expression with exact value $\frac{1}{2}$.

Rewrite tangent as sine divided by cosine.

Tangent is not a ratio in its own right: it is sine over cosine. So tangent has no value wherever the cosine is zero.

Test the angle where cosine is zero.

At $90^\circ$ the adjacent side has shrunk to nothing, so $\cos 90^\circ = 0$ and the tangent would need a division by zero. Division by zero has no meaning, so $\tan 90^\circ$ is undefined.

Check every other option does have a value.

These are ordinary entries of the exact table, so they are all perfectly well defined.

Check the last two options as well.

They too have exact values, so the undefined one has to be the remaining option.

Derive the 45 degree values from a right-angled isosceles triangle.

Take a right-angled triangle with two sides of $1$. Its two other angles are equal, so each is $45^\circ$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. So $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45^\circ = \frac{1}{1} = 1$.

Select the undefined expression.

Only $\tan 90^\circ$ is undefined, because it would need a division by zero.

Write down the exact value of each expression.

Ordering is impossible until every expression has been turned into an exact number.

Put the exact values in ascending order.

Comparing the surds by squaring them, rather than by rounding them, gives this ordering exactly.

Read the matching order of the original expressions.

Replacing each exact value by the expression it came from gives the list in the right order.

Sense-check with the sizes of the angles.

Sine increases and cosine decreases as the angle grows, and tangent overtakes both once the angle passes $45^\circ$. The ordering fits.

Derive the 30 and 60 degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Derive the 45 degree values from a right-angled isosceles triangle.

Take a right-angled triangle with two sides of $1$. Its two other angles are equal, so each is $45^\circ$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. So $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45^\circ = \frac{1}{1} = 1$.

Recall the three trigonometric ratios.

SOH CAH TOA: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent. $O$ and $A$ are measured from the angle you are working with.

Recall the exact values of sine.

Sine climbs steadily from $0$ at $0^\circ$ to $1$ at $90^\circ$. The numerators run $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ over a denominator of $2$, which is the quickest way to remember the row.

Recall the exact values of cosine.

Cosine is the sine row read backwards: it falls from $1$ at $0^\circ$ to $0$ at $90^\circ$. That is because $\cos \theta = \sin(90^\circ - \theta)$.

Select the list in ascending order.

This is the only option whose values increase from left to right.

Write down the exact value of every option.

Nothing can be compared until every option is an exact number, so start by evaluating all five.

Write down the exact value of the remaining options.

These two complete the list of five exact values.

Compare the exact values by squaring them.

Two surds are compared exactly by squaring: the bigger square belongs to the bigger positive number, so no decimals are needed at any point.

Sense-check against how the ratios behave.

Sine grows and cosine shrinks as the angle grows from $0^\circ$ to $90^\circ$, and tangent grows fastest of all. The ordering found above agrees with that.

Derive the 30 and 60 degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Derive the 45 degree values from a right-angled isosceles triangle.

Take a right-angled triangle with two sides of $1$. Its two other angles are equal, so each is $45^\circ$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. So $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45^\circ = \frac{1}{1} = 1$.

Recall the three trigonometric ratios.

SOH CAH TOA: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent. $O$ and $A$ are measured from the angle you are working with.

Recall the exact values of sine.

Sine climbs steadily from $0$ at $0^\circ$ to $1$ at $90^\circ$. The numerators run $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ over a denominator of $2$, which is the quickest way to remember the row.

Recall the exact values of cosine.

Cosine is the sine row read backwards: it falls from $1$ at $0^\circ$ to $0$ at $90^\circ$. That is because $\cos \theta = \sin(90^\circ - \theta)$.

Recall the exact values of tangent.

Tangent is sine divided by cosine. It is only asked for at $0^\circ$, $30^\circ$, $45^\circ$ and $60^\circ$, because at $90^\circ$ the cosine is $0$ and $\tan 90^\circ$ is undefined.

Note that tangent is sine divided by cosine.

Every exact tangent can be rebuilt from the sine and cosine of the same angle: for example $\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\sqrt{3}}{2} \div \frac{1}{2} = \sqrt{3}$.

Note the identity that ties sine and cosine together.

The two shorter sides and the hypotenuse obey Pythagoras, so for any angle $\sin^2 \theta + \cos^2 \theta = 1$. At $30^\circ$ this reads $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$, which is a useful check.

Note the link between an angle and its complement.

The side opposite one acute angle is the side adjacent to the other, so $\sin 30^\circ = \cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$. Half the table is the other half.

Recall the surd rules used here.

Multiplying a surd by itself removes the root altogether, and two different surds multiply into a single root: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$.

Select the correct option.

$\cos 90^\circ = 0$, which is the least of the five exact values.