Exact values at a known angle

Find an unknown constant

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

How it is set in the exam

Paper 1 (no calculator), 2–4 marks a part. Rare on its own (1 of 20 exam versions: two "show that" parts of a Section B question, 6 marks between them), but the same exact-value step sits inside many other types (Where it hides).

Spot it in the exam

Exam phrases to spot (3)

“given that $f\left(\frac{\pi}{6}\right) = \ldots$ , show that $k = \ldots$ ”
“find the exact value of the constant”
“without a calculator”

In disguise (2)

the value is a point on the graph at a standard angle, such as $\left(\frac{\pi}{4}, 3\right)$
the formula is a model in a context (a force, a length, a time), and the angle is where a measurement was made (the formula is neither a sinusoid nor a tan model)

Variants (2)

a constant from the value at $0$ , where $\sin 0 = 0$ and $\cos 0 = 1$ remove a term (a "show that")
a second constant from the value at $\frac{\pi}{6}$ , $\frac{\pi}{4}$ or $\frac{\pi}{3}$ , ending in a surd that has to be rationalised (a "show that")

Looks like it, but isn't (4)

the formula is a sinusoidal model and its value at a time is asked: Read a sinusoid's features
the formula is a sinusoid, $a\sin(b(x + c)) + d$ or $a\cos(b(x + c)) + d$ , or a tan model, $a\tan(bx) + d$ , and its constants come from its graph or its greatest and least values, or two unknown constants come from two given points, solved together: Find a trig model's equation
only a ratio of an unknown angle is known: Exact values without the angle: from one ratio to another
the value is given and the angle is unknown: Solve a trig equation on an interval

Worked example

A constant from $f(0)$ , then a surd from $f\left(\frac{\pi}{3}\right)$

Paper 1 (no calculator)

Question

Let $f(x) = \dfrac{a + b\sin x}{\cos x}$ , for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , where $a$ and $b$ are constants. The graph of $f$ passes through the points $(0, 3)$ and $\left(\frac{\pi}{3}, 12\right)$ .
(a) Show that $a = 3$ . [2]
(b) Show that $b = 2\sqrt{3}$ . [4]

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Steps

Substitute
the angle and the given value into the formula, and write the whole equation.
Exact values:
replace each $\sin$ , $\cos$ and $\tan$ by its exact value, with the quadrant's sign (Exact values). Work out a compound argument first: at $x = \frac{\pi}{6}$ , $\sin 2x = \sin\frac{\pi}{3}$ .
Clear the fractions
, including a denominator of the formula itself: you now have a linear equation in the constant. With two constants and no value at $0$ to remove one, you get two linear equations: solve them simultaneously.
Solve and simplify:
rationalise a surd denominator, e.g. $\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$ .
In a "show that",
the line with the exact values substituted must be seen, then each step to the given value. Substituting the given answer to check it is working backwards, and scores nothing.

Common mistakes

Decimals in a "show that"
$\sin\frac{\pi}{3} = 0.866\ldots$ can never reach an exact surd. Use $\frac{\sqrt{3}}{2}$ .
The argument worked out wrongly
at $x = \frac{\pi}{6}$ , $\sin 2x = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ , not $2\sin\frac{\pi}{6} = 1$ .
Substituting into the numerator only
and forgetting the formula's denominator.

More mistakes (3)

The wrong sign outside the first quadrant
$\cos\frac{2\pi}{3} = -\frac{1}{2}$ , not $\frac{1}{2}$ .
Working backwards
putting $b = 2\sqrt{3}$ into the formula and checking that it gives $12$ proves nothing and earns nothing.
Stopping short of the given form
the last mark is for the simplified linear equation, but a "show that" must visibly reach the given answer: write $\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$ .

Easy to confuse with

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