🎯What you need to be able to do
- Use \( \sec\theta \), \( \csc\theta \) and \( \cot\theta \), and sketch their graphs.
- Use the identities \( 1 + \tan^{2}\theta = \sec^{2}\theta \) and \( 1 + \cot^{2}\theta = \csc^{2}\theta \).
- Use the compound angle formulae for sine, cosine and tangent.
- Use the double angle formulae, including all three forms of \( \cos 2\theta \).
- Express \( a\sin\theta + b\cos\theta \) in the form \( R\sin(\theta \pm \alpha) \) or \( R\cos(\theta \pm \alpha) \).
- Use that form to solve equations and to find maximum and minimum values.
- Prove identities involving any of the above.
📚The mathematics
The reciprocal functions
\( \sec\theta = \dfrac{1}{\cos\theta} \)
\( \csc\theta = \dfrac{1}{\sin\theta} \)
\( \cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta} \)
Note the mismatch that catches everyone: secant pairs with cosine, not sine. The
third letter of each name is the giveaway — secant with cosine, cosecant
with sine.
Each is undefined wherever its partner is zero, so \( \sec\theta \) has vertical asymptotes at
\( 90^{\circ}, 270^{\circ}, \dots \) and never takes a value between \( -1 \) and 1.
The Pythagorean identities
\( \sin^{2}\theta + \cos^{2}\theta = 1 \)
\( 1 + \tan^{2}\theta = \sec^{2}\theta \)
\( 1 + \cot^{2}\theta = \csc^{2}\theta \)
The second and third are not separate facts to memorise: divide the first by
\( \cos^{2}\theta \) to get the second, and by \( \sin^{2}\theta \) to get the third. Deriving them
once means never mixing them up.
Compound angle formulae
\( \sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \)
\( \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \)
\( \tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \)
These are given in the list of formulae, so the marks are for using them, not recalling them.
Watch the signs: in the cosine and tangent formulae the sign on the right flips relative to
the left. The sine formula keeps the same sign, which is why it feels easier and why the cosine one
is misquoted more often.
Double angle formulae
\( \sin 2\theta = 2\sin\theta\cos\theta \)
\( \cos 2\theta = \cos^{2}\theta - \sin^{2}\theta \)
\( = 2\cos^{2}\theta - 1 \)
\( = 1 - 2\sin^{2}\theta \)
\( \tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^{2}\theta} \)
Set \( B = A \) in the compound formulae and all of these fall out, including the three forms of
\( \cos 2\theta \) once you substitute the Pythagorean identity.
Three versions of \( \cos 2\theta \) exist so you can pick the one that leaves only the function
you already have. If an equation contains \( \cos 2\theta \) and \( \sin\theta \), use
\( 1 - 2\sin^{2}\theta \) and everything becomes a quadratic in \( \sin\theta \). Choosing the right
form is most of the skill.
Rearranged, the last two also give the power-reduction forms that make
\( \cos^{2}\theta \) and \( \sin^{2}\theta \) integrable in
P2/3 5:
\( \cos^{2}\theta = \tfrac{1}{2}\left(1 + \cos 2\theta\right) \)
\( \sin^{2}\theta = \tfrac{1}{2}\left(1 - \cos 2\theta\right) \)
The \( R \) form
Any expression \( a\sin\theta + b\cos\theta \) can be written as a single trigonometric function:
\[ a\sin\theta + b\cos\theta = R\sin(\theta + \alpha),
\qquad R = \sqrt{a^{2}+b^{2}}, \quad \tan\alpha = \frac{b}{a} \]
The method is always the same: expand the target form with the compound angle formula, compare
coefficients of \( \sin\theta \) and \( \cos\theta \), then get \(R\) by squaring and adding and
\( \alpha \) by dividing. Take \(R\) positive and \( \alpha \) acute unless told otherwise.
Why bother? Because the single-function form makes two things immediate:
- Solving equations. \( 3\sin\theta + 4\cos\theta = 2 \) is intractable as it
stands; as \( 5\sin(\theta + \alpha) = 2 \) it is routine.
- Maxima and minima. Since \( \sin \) ranges over \( [-1, 1] \), the maximum of
\( R\sin(\theta+\alpha) \) is \(R\) and the minimum is \( -R \) — and you can say
where they occur by setting the bracket to \( 90^{\circ} \) or \( 270^{\circ} \). No
calculus required, which is the point.
Match the form the question asks for. \( R\sin(\theta - \alpha) \),
\( R\cos(\theta + \alpha) \) and the rest all work, but they give
different values of
\( \alpha \). Expand the exact form requested and compare coefficients with that — do not
derive one form and hope it matches. And when solving, remember the interval must be shifted by
\( \alpha \) before you find all the solutions, exactly as in
P1 5.
✏️Worked example
(a) Express \( 7\sin\theta - 24\cos\theta \) in the form \( R\sin(\theta - \alpha) \), where
\( R > 0 \) and \( 0^{\circ} < \alpha < 90^{\circ} \).
(b) Hence solve \( 7\sin\theta - 24\cos\theta = 15 \) for \( 0^{\circ} \le \theta \le 360^{\circ} \).
(c) State the greatest value of \( 7\sin\theta - 24\cos\theta + 30 \) and the value of
\( \theta \) at which it occurs.
(a) Expand the required form:
\( R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha \). Comparing
coefficients:
\( R\cos\alpha = 7 \)
\( R\sin\alpha = 24 \)
Squaring and adding: \( R^{2} = 49 + 576 = 625 \), so \( R = 25 \). Dividing:
\( \tan\alpha = \dfrac{24}{7} \), so \( \alpha = 73.74^{\circ} \). Hence
\[ 7\sin\theta - 24\cos\theta = 25\sin\left(\theta - 73.74^{\circ}\right) \]
(b) The equation becomes \( 25\sin(\theta - 73.74^{\circ}) = 15 \), so
\( \sin(\theta - 73.74^{\circ}) = 0.6 \).
Let \( u = \theta - 73.74^{\circ} \) and shift the interval: as \( \theta \) runs from
\( 0^{\circ} \) to \( 360^{\circ} \), \(u\) runs from \( -73.74^{\circ} \) to
\( 286.26^{\circ} \).
From \( \sin u = 0.6 \), the principal value is \( u = 36.87^{\circ} \), and the second-quadrant
solution is \( 180^{\circ} - 36.87^{\circ} = 143.13^{\circ} \). Both lie in the interval. Adding
\( 360^{\circ} \) to either takes us outside it.
Converting back, \( \theta = u + 73.74^{\circ} \):
\[ \theta = 110.6^{\circ} \quad\text{or}\quad \theta = 216.9^{\circ} \]
(c) The maximum of \( 25\sin(\theta - 73.74^{\circ}) \) is 25, so the greatest
value of the whole expression is \( 25 + 30 = 55 \). It occurs when the bracket equals
\( 90^{\circ} \), that is \( \theta - 73.74^{\circ} = 90^{\circ} \), giving
\( \theta = 163.7^{\circ} \).
Check it. Substitute \( \theta = 110.6^{\circ} \) into the original:
\( 7\sin 110.6^{\circ} - 24\cos 110.6^{\circ} = 7(0.9362) - 24(-0.3516) = 6.55 + 8.44 = 15.0 \). ✓
And in (c), \( 7\sin 163.7^{\circ} - 24\cos 163.7^{\circ} = 7(0.2800) - 24(-0.9600) = 1.96 + 23.04
= 25 \), which is exactly \(R\) as it should be. ✓
Shift the interval before solving, not after. Working in \( \theta \) directly
and then subtracting \( \alpha \) at the end loses solutions, because the interval for \(u\) is
displaced — here it starts at a negative angle, which is exactly the region
candidates forget to consider. Write the transformed interval down explicitly before touching
the calculator.
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. Solve \( 3\sec^{2}\theta - 5\tan\theta - 4 = 0 \) for \( 0^{\circ} \le \theta \le 180^{\circ} \).
Use \( \sec^{2}\theta = 1 + \tan^{2}\theta \): \( 3\left(1 + \tan^{2}\theta\right) - 5\tan\theta - 4 = 0 \), so \( 3\tan^{2}\theta - 5\tan\theta - 1 = 0 \). By the quadratic formula, \( \tan\theta = \dfrac{5 \pm \sqrt{25 + 12}}{6} = \dfrac{5 \pm 6.083}{6} \), giving \( \tan\theta = 1.847 \) or \( \tan\theta = -0.1805 \). From the first, \( \theta = 61.6^{\circ} \). From the second, the principal value is \( -10.2^{\circ} \), so add \( 180^{\circ} \) to land in the interval: \( \theta = 169.8^{\circ} \).
2. Prove that \( \dfrac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta \).
Work on the left-hand side, choosing the form of \( \cos 2\theta \) that makes the denominator collapse: \( 1 + \cos 2\theta = 1 + \left(2\cos^{2}\theta - 1\right) = 2\cos^{2}\theta \). The numerator is \( 2\sin\theta\cos\theta \). So LHS \( = \dfrac{2\sin\theta\cos\theta}{2\cos^{2}\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta \equiv \) RHS, as required. Picking \( 2\cos^{2}\theta - 1 \) rather than one of the other two forms is the whole trick — it is the one that leaves no stray constant.
3. Solve \( \cos 2\theta + 3\sin\theta = 2 \) for \( 0^{\circ} \le \theta \le 360^{\circ} \).
The equation mixes \( \cos 2\theta \) with \( \sin\theta \), so use \( \cos 2\theta = 1 - 2\sin^{2}\theta \): \( 1 - 2\sin^{2}\theta + 3\sin\theta = 2 \), giving \( 2\sin^{2}\theta - 3\sin\theta + 1 = 0 \) and \( (2\sin\theta - 1)(\sin\theta - 1) = 0 \). So \( \sin\theta = \tfrac12 \) or \( \sin\theta = 1 \). From the first, \( \theta = 30^{\circ} \) or \( 150^{\circ} \); from the second, \( \theta = 90^{\circ} \).
4. Express \( 5\cos\theta + 12\sin\theta \) in the form \( R\cos(\theta - \alpha) \) with \( R > 0 \) and \( \alpha \) acute.
Expand: \( R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha \). Comparing, \( R\cos\alpha = 5 \) and \( R\sin\alpha = 12 \). So \( R = \sqrt{25 + 144} = 13 \), and \( \tan\alpha = \dfrac{12}{5} \), giving \( \alpha = 67.4^{\circ} \). Hence \( 13\cos\left(\theta - 67.4^{\circ}\right) \). Note the cosine form pairs \( \cos\theta \) with \( \cos\alpha \) — a different pairing from the sine form, which is why the form matters.
5. Given \( \sin A = \tfrac{3}{5} \) with \(A\) acute, and \( \cos B = \tfrac{5}{13} \) with \(B\) acute, find \( \sin(A + B) \) exactly.
Build the missing ratios with Pythagoras. For \(A\): \( \cos A = \tfrac45 \) (positive, since \(A\) is acute). For \(B\): \( \sin B = \tfrac{12}{13} \). Then \( \sin(A+B) = \sin A\cos B + \cos A\sin B = \left(\tfrac35\right)\left(\tfrac{5}{13}\right) + \left(\tfrac45\right)\left(\tfrac{12}{13}\right) = \tfrac{15}{65} + \tfrac{48}{65} = \tfrac{63}{65} \).
6. Find the maximum and minimum values of \( \dfrac{10}{4 + 3\sin\theta + 4\cos\theta} \).
First write \( 3\sin\theta + 4\cos\theta = R\sin(\theta + \alpha) \) with \( R = \sqrt{9+16} = 5 \). So the denominator is \( 4 + 5\sin(\theta+\alpha) \), which ranges from \( 4 - 5 = -1 \) to \( 4 + 5 = 9 \). The fraction is greatest when the denominator is smallest positive and least when it is largest — but note the denominator passes through zero, so the expression is unbounded. Restricting to where the denominator is positive: as it approaches 0 the fraction grows without limit, and its smallest value is \( \dfrac{10}{9} \) when the denominator is 9, at \( \sin(\theta+\alpha) = 1 \). This is a good reminder to check whether a denominator can vanish before quoting a maximum.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and
everything above it on this page still stands.
- Desmos — plot \( a\sin\theta + b\cos\theta \) and \( R\sin(\theta+\alpha) \) together and watch them coincide
- GeoGebra — the unit circle, for the compound angle geometry
- Khan Academy — trigonometric identities and the double angle formulae