Hard GCSE Exact trigonometric values Questions

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.

Say what rationalising the denominator means.

An answer is not in its final form while a surd sits in the denominator. The value must not change, only the way it is written.

Multiply the top and the bottom by the surd in the denominator.

Multiplying top and bottom by $\sqrt{3}$ is multiplying by $1$, so the value is untouched, and $\sqrt{3} \times \sqrt{3} = 3$ clears the surd from the bottom.

Check the value is unchanged by squaring both forms.

Both the original and the rationalised form square to the same number, so they are the same value written two ways.

Reject the line that changes the value.

This line does not multiply the top and the bottom by the same thing, so it changes the value of the fraction. That makes it wrong however tidy it looks.

Reject the line whose last equals sign is false.

This line sets up the multiplication correctly but then writes down the wrong denominator, so its final equals sign is simply untrue.

Check the surd by squaring it.

Squaring $\frac{\sqrt{3}}{3}$ gives $\frac{1}{3}$, which is a whole number or a simple fraction. That is the sign of a correctly simplified surd.

Recall how to rationalise a denominator.

A surd is never left on the bottom of a fraction. Multiply top and bottom by that surd: since $\sqrt{n} \times \sqrt{n} = n$, the denominator turns into a whole number and the value of the fraction is unchanged.

Recall the surd rules used here.

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

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.

Select the correct line of working.

Every equals sign holds and the answer ends as $\frac{\sqrt{3}}{3}$, with a rational denominator.

Read the sides of the special triangle off the diagram.

The triangle in the question is drawn true to scale, and every exact value comes from the ratios of these three sides.

Write the required ratio using those sides.

Reading the correct pair of sides off the triangle, and then clearing the surd from the denominator, gives $\frac{\sqrt{2}}{2}$.

Check the final value is in its simplest form.

The answer $\frac{\sqrt{2}}{2}$ has no surd in its denominator and no common factor left to cancel, so it is fully simplified.

Reject the line that picks the wrong pair of sides.

This line uses the wrong two sides of the triangle, so its very first step is already false.

Reject the line that fails to rationalise correctly.

This line goes wrong at the last equals sign: the two sides of it are simply not the same number.

Check the surd by squaring it.

Squaring $\frac{\sqrt{2}}{2}$ gives $\frac{1}{2}$, which is a whole number or a simple fraction. That is the sign of a correctly simplified surd.

Recall how to rationalise a denominator.

A surd is never left on the bottom of a fraction. Multiply top and bottom by that surd: since $\sqrt{n} \times \sqrt{n} = n$, the denominator turns into a whole number and the value of the fraction is unchanged.

Recall the surd rules used here.

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

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.

Select the correct line of working.

Every equals sign in this line is true, and it ends at the fully simplified exact value $\frac{\sqrt{2}}{2}$.

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 down the exact value of cos 30 degrees.

From the exact table, $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

Write down the exact value of tan 60 degrees.

From the exact table, $\tan 60^\circ = \sqrt{3}$.

Substitute the exact value of every trig ratio.

Replacing each ratio by its exact value turns the expression into $\frac{2}{\frac{\sqrt{3}}{2}} + \sqrt{3}$, which is now just surd arithmetic.

Work out the arithmetic exactly.

Combining the exact values gives $\frac{7\sqrt{3}}{3}$. Nothing is rounded at any stage, so the answer is exact.

Check the surd by squaring it.

Squaring $\frac{7\sqrt{3}}{3}$ gives $\frac{49}{3}$, which is a whole number or a simple fraction. That is the sign of a correctly simplified surd.

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.

State the exact value.

The exact value is $\frac{7\sqrt{3}}{3}$, in simplest form with a rational denominator.

Split the equilateral triangle down the middle.

Every angle of an equilateral triangle is $60^\circ$. Dropping the perpendicular from the top vertex cuts it into two identical right-angled triangles, each with a $30^\circ$, a $60^\circ$ and a $90^\circ$ angle.

Pick out the right-angled triangle to work in.

One half has a hypotenuse of $7$ cm (a full side), a base of $\frac{7}{2}$ cm (half a side) and the height $h$ opposite the $60^\circ$ angle at the base.

Choose the ratio linking the height to the side.

The height is opposite the $60^\circ$ angle and the full side is the hypotenuse, so sine is the ratio to use.

Write down the exact value of sin 60 degrees.

From the exact table, $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

Work out the exact height.

The exact height is $\frac{7\sqrt{3}}{2}$ cm. It is left as a surd, not rounded.

Check the units.

Every length in the question is in cm, so the height is measured in cm.

Check the surd by squaring it.

Squaring $\frac{7\sqrt{3}}{2}$ gives $\frac{147}{4}$, which is a whole number or a simple fraction. That is the sign of a correctly simplified surd.

Check the height is shorter than a side.

The height is a leg of a right-angled triangle whose hypotenuse is a full side, so it must come out shorter than $7$ cm. It does.

Check the three sides with Pythagoras.

The squares of the two shorter sides add to $49$, which is exactly the square of the hypotenuse. The three exact lengths fit Pythagoras, so they really are the sides of this triangle.

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 $\cos \theta = \sin(90^\circ - \theta)$.

State the exact height of the triangle.

The exact height is $\frac{7\sqrt{3}}{2}$ cm.