Exact values without the angle

From one ratio to another

In 4 of 20 recent exams
Paper 1, no calculator
2 to 6 marks

How it is set in the exam

Paper 1 (no calculator) only, 2–6 marks a part; typical (4 of 20 exam versions), often filling a whole 6–9-mark question. It is usually set in a triangle: the sine rule or the cosine rule gives one ratio, and an identity gives the one asked for. When the part asks for an area or a volume, this step belongs to Area of a triangle or Volume and surface area instead.

Spot it in the exam

Exam phrases to spot (6)

“given that $\cos\theta = \ldots$ , find the exact value of $\sin\theta$ ”
“… of $\sin 2\theta$ ”
“ $\theta$ is acute”
“ $\theta$ is obtuse”
“show that $\cos\theta = \ldots$ ”
“find the exact value of the sine of”

In disguise (3)

the first ratio comes out of a triangle: the cosine rule gives $\cos 2\theta$ from three sides, or the sine rule with the sides opposite $\theta$ and $2\theta$ gives $\cos\theta$
the size of the angle is given by an interval such as $\frac{\pi}{2} < \theta < \pi$ , or by the figure ("an obtuse angle", "half of an angle of the triangle"), and it decides the sign
an angle on a straight line with a known angle: the same sine, the opposite cosine

Variants (5)

$\cos\theta \to \sin\theta$ (acute), then $\sin 2\theta$ or $\cos 2\theta$
$\sin 2\theta$ reused for the supplementary angle
the cosine rule gives $\cos 2\theta$ , then $\cos\theta$ (positive root, in a requested surd form)
the sine rule with $\theta$ and $2\theta$ gives $\cos\theta$ (cancel $\sin\theta$ )
$\tan\theta \to \sin\theta$ and $\cos\theta$ from a right triangle

Looks like it, but isn't (5)

an area or a volume is asked, and the ratio is one step on the way: Area of a triangle, Volume and surface area
the angle is known: Exact values at a known angle
"solve", or "find $\theta$ ": every angle in an interval is wanted: Solve a trig equation on an interval
a condition on a solid or a figure (two areas, or two lengths, equal) gives an equation in $\sin\theta$ and $\cos\theta$ to rearrange into " $\tan\theta = \ldots$ ", even when no angle is asked: Solve a trig equation on an interval. The sine rule with angles $\theta$ and $2\theta$ (Example 2.2) stays here.
a side of a triangle is asked, and the given cosine goes straight into the cosine rule, with no identity: Find a side or an angle: choose the rule

Worked example

From $\cos\theta$ to $\sin\theta$ , $\sin 2\theta$ and $\cos 2\theta$

Paper 1 (no calculator)

Question

The angle $\theta$ is acute, and $\cos\theta = \frac{2}{5}$ . Find the exact value of
(a) $\sin\theta$ [2]
(b) $\sin 2\theta$ [2]
(c) $\cos 2\theta$ . [2]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

Which method?

$\cos\theta$ or $\sin\theta$
Asked for
the other one
Tool
$\sin^2\theta + \cos^2\theta = 1$ or a right triangle; the sign from the quadrant
$\sin\theta$ and $\cos\theta$
Asked for
$\sin 2\theta$ or $\cos 2\theta$
Tool
the double angle; for $\cos 2\theta$ , the form that uses what you already have
$\cos 2\theta$
Asked for
$\cos\theta$ or $\sin\theta$
Tool
$\cos 2\theta = 2\cos^2\theta - 1$ (or $1 - 2\sin^2\theta$ ), then the square root with the right sign
$\tan\theta$
Asked for
$\sin\theta$ and $\cos\theta$
Tool
a right triangle: opposite and adjacent from $\tan\theta$ , the hypotenuse by Pythagoras; signs from the quadrant
angles $\theta$ and $2\theta$ in one triangle, and the sides opposite them
Asked for
$\cos\theta$
Tool
the sine rule, $\sin 2\theta = 2\sin\theta\cos\theta$ , then cancel $\sin\theta$ (never $0$ in a triangle)
an angle supplementary to a known one
Asked for
its sine or cosine
Tool
$\sin(\pi - x) = \sin x$ , $\cos(\pi - x) = -\cos x$

Given	Asked for	Tool
$\cos\theta$ or $\sin\theta$	the other one	$\sin^2\theta + \cos^2\theta = 1$ or a right triangle; the sign from the quadrant
$\sin\theta$ and $\cos\theta$	$\sin 2\theta$ or $\cos 2\theta$	the double angle; for $\cos 2\theta$ , the form that uses what you already have
$\cos 2\theta$	$\cos\theta$ or $\sin\theta$	$\cos 2\theta = 2\cos^2\theta - 1$ (or $1 - 2\sin^2\theta$ ), then the square root with the right sign
$\tan\theta$	$\sin\theta$ and $\cos\theta$	a right triangle: opposite and adjacent from $\tan\theta$ , the hypotenuse by Pythagoras; signs from the quadrant
angles $\theta$ and $2\theta$ in one triangle, and the sides opposite them	$\cos\theta$	the sine rule, $\sin 2\theta = 2\sin\theta\cos\theta$ , then cancel $\sin\theta$ (never $0$ in a triangle)
an angle supplementary to a known one	its sine or cosine	$\sin(\pi - x) = \sin x$ , $\cos(\pi - x) = -\cos x$

Steps

The sign first:
from the words (acute, obtuse, an interval, an angle of a triangle), decide the quadrant (an angle of a triangle only tells you its sine is positive), and so the sign of the answer.
Choose the tool
from the table.
Substitute exactly
and simplify: $\sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$ ; rationalise if the requested form needs it.
Check:
your sine and cosine of the same angle satisfy $\sin^2 + \cos^2 = 1$ , and the sign agrees with step 1.
Pass the value on
(to an area, the sine rule or a volume), exactly.

Common mistakes

The sign by habit
a square root is positive; the quadrant decides the sign. For an obtuse angle, $\cos\theta$ and $\tan\theta$ are negative.
Both roots left in
when the sign is known: for an acute angle (including half of an angle of a triangle) every ratio is positive, and the sine of any angle of a triangle is positive, so only the positive value may appear in the final answer (markschemes note that the final answer must be positive). The cosine or tangent of an angle of a triangle can still be negative (an obtuse angle). Write the reason in words: it is what makes the sign right.
$\sin 2\theta = 2\sin\theta$ ,
or $\cos 2\theta = 2\cos\theta - 1$ with the square forgotten.

More mistakes (4)

Finding $\theta$ first
on a calculator: impossible on , and a decimal angle can't give an exact answer.
The requested form not reached
$\sqrt{\frac{3}{8}}$ is not $\frac{\sqrt{p}}{q}$ with whole numbers; missing the rationalising step loses the final mark.
A "show that" with the key line missing
in Example 2.2 the line with $2\sin\theta\cos\theta$ , and the cancelling, must be seen before the given answer. In a surd "show that" (from $\cos\theta = \frac{3}{7}$ to a given $\sin\theta = \frac{2\sqrt{10}}{7}$ ), the mark is for $\sin^2\theta = \frac{40}{49}$ , but the given form must still be reached visibly: $\sqrt{\frac{40}{49}} = \frac{\sqrt{40}}{7} = \frac{2\sqrt{10}}{7}$ .
Putting the cosine into $\frac{1}{2}ab\sin C$
when an area follows: the formula needs the sine you found (Area of a triangle).

Easy to confuse with

Next typeExact values at a known angle

Now try one like it

Keep going

Easy to confuse with