Further trigonometry – Year 13 A level maths revision notes to download as PDF

Worked example 1

Find an approximation for \(\dfrac{1-\cos2\theta}{\theta\tan3\theta}\) when \(\theta\) is small.

\(1-\cos2\theta\approx\dfrac{(2\theta)^2}{2}=2\theta^2\) and \(\theta\tan3\theta\approx3\theta^2\), so the ratio is approximately \(\tfrac23\).

Worked example 2

Solve \(\sec x=3\) and \(\operatorname{cosec}^2x-\cot x-3=0\) for \(0^\circ\le x<360^\circ\).

\(\cos x=\tfrac13\), so \(x=70.5^\circ\) or \(289.5^\circ\). For the second, replace \(\operatorname{cosec}^2x\) by \(1+\cot^2x\): \(\cot^2x-\cot x-2=0\), so \(\cot x=2\) or \(-1\), i.e. \(\tan x=\tfrac12\) or \(-1\), giving \(x=26.6^\circ,206.6^\circ,135^\circ,315^\circ\).

Worked example 3

Find the exact value of \(\sin75^\circ\).

\(\sin(45^\circ+30^\circ)=\dfrac{\sqrt2}2\cdot\dfrac{\sqrt3}2+\dfrac{\sqrt2}2\cdot\dfrac12=\dfrac{\sqrt6+\sqrt2}4\).

Worked example 4

\(\sin A=\tfrac35\) and \(\cos B=\tfrac5{13}\) with both angles acute. Find \(\sin(A+B)\).

\(\cos A=\tfrac45\), \(\sin B=\tfrac{12}{13}\), so \(\sin(A+B)=\tfrac35\cdot\tfrac5{13}+\tfrac45\cdot\tfrac{12}{13}=\tfrac{63}{65}\).

Worked example 5

Solve \(\cos2x=\sin x\) for \(0^\circ\le x<360^\circ\).

\(1-2\sin^2x=\sin x\), so \(2\sin^2x+\sin x-1=0\), \((2\sin x-1)(\sin x+1)=0\), and \(x=30^\circ,150^\circ,270^\circ\).

Worked example 6

Prove that \(\dfrac{\sin2x}{1+\cos2x}\equiv\tan x\).

\(\dfrac{2\sin x\cos x}{1+(2\cos^2x-1)}=\dfrac{2\sin x\cos x}{2\cos^2x}=\tan x\). \(\blacksquare\)

Worked example 7

Write \(3\sin x+4\cos x\) in the form \(R\sin(x+\alpha)\) and solve \(3\sin\theta+4\cos\theta=2\) for \(0^\circ\le\theta<360^\circ\).

\(R=5\), \(\tan\alpha=\tfrac43\), so \(\alpha=53.13^\circ\) and the expression is \(5\sin(x+53.13^\circ)\), with maximum 5 at \(x=36.87^\circ\). Then \(\sin(\theta+53.13^\circ)=0.4\), so \(\theta+53.13^\circ=23.58^\circ\) or \(156.42^\circ\) (add \(360^\circ\) if needed): \(\theta=103.3^\circ\) or \(330.4^\circ\). The expression \(10+3\sin x+4\cos x\) lies between 5 and 15.

Worked example 8

Prove that \(\cos3x\equiv4\cos^3x-3\cos x\).

\(\cos3x=\cos(2x+x)=\cos2x\cos x-\sin2x\sin x=(2\cos^2x-1)\cos x-2\sin^2x\cos x=2\cos^3x-\cos x-2(1-\cos^2x)\cos x=4\cos^3x-3\cos x\). \(\blacksquare\)

Worked example 9: a limit by small angles

Find an approximation for \(\dfrac{\tan5\theta\sin2\theta}{\theta(1-\cos\theta)}\) for small \(\theta\) and explain what happens as \(\theta\to0\).

\(\tan5\theta\approx5\theta\), \(\sin2\theta\approx2\theta\) and \(1-\cos\theta\approx\dfrac{\theta^2}2\). The expression is approximately \(\dfrac{5\theta\cdot2\theta}{\theta\cdot\theta^2/2}=\dfrac{10\theta^2}{\theta^3/2}=\dfrac{20}{\theta}\), which grows without bound as \(\theta\to0\): there is no finite limit.

Worked example 10: proving with reciprocal functions

Prove that \(\sec x+\tan x\equiv\dfrac{1+\sin x}{\cos x}\) and hence that \(\dfrac1{\sec x-\tan x}\equiv\sec x+\tan x\).

\(\sec x+\tan x=\dfrac1{\cos x}+\dfrac{\sin x}{\cos x}=\dfrac{1+\sin x}{\cos x}\). For the second part, \((\sec x-\tan x)(\sec x+\tan x)=\sec^2x-\tan^2x=1\), so \(\dfrac1{\sec x-\tan x}=\sec x+\tan x\). \(\blacksquare\)

Worked example 11: a double angle identity

Prove that \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\).

\(1-\cos2x=2\sin^2x\) and \(\sin2x=2\sin x\cos x\), so the expression is \(\dfrac{2\sin^2x}{2\sin x\cos x}=\dfrac{\sin x}{\cos x}=\tan x\). \(\blacksquare\)

Worked example 12: exact value with tan

Given that \(\tan A=\dfrac34\) and \(A\) is acute, find \(\sin2A\), \(\cos2A\) and \(\tan2A\).

The triangle is 3, 4, 5, so \(\sin A=\dfrac35\) and \(\cos A=\dfrac45\). \(\sin2A=2\cdot\dfrac35\cdot\dfrac45=\dfrac{24}{25}\). \(\cos2A=\dfrac{16}{25}-\dfrac9{25}=\dfrac7{25}\). \(\tan2A=\dfrac{24}7\). Check with the formula \(\dfrac{2\tan A}{1-\tan^2A}=\dfrac{3/2}{7/16}=\dfrac{24}7\).

Worked example 13: R form with the cosine

Write \(\sin x-\sqrt3\cos x\) in the form \(R\sin(x-\alpha)\) and solve \(\sin x-\sqrt3\cos x=1\) for \(0^\circ\le x<360^\circ\).

\(R=\sqrt{1+3}=2\) and \(\tan\alpha=\sqrt3\), so \(\alpha=60^\circ\) and the expression is \(2\sin(x-60^\circ)\). Then \(\sin(x-60^\circ)=\tfrac12\), so \(x-60^\circ=30^\circ\) or \(150^\circ\), giving \(x=90^\circ\) or \(210^\circ\).

Worked example 14: range of a function with the R form

Find the greatest and least values of \(\dfrac{12}{5+3\sin x+4\cos x}\).

The denominator is \(5+5\sin(x+53.13^\circ)\), which lies between 0 and 10. The value 0 is reached when the sine is \(-1\), so the expression is not defined there. Close to that point it becomes very large, so there is no greatest value; the least value is \(\dfrac{12}{10}=1.2\) when the sine is \(+1\).

Worked example 15: an equation with a double argument

Solve \(\sin2x=0.6\) for \(0^\circ\le x<360^\circ\).

Let \(u=2x\), which ranges over \(0^\circ\le u<720^\circ\). \(\sin u=0.6\) gives \(u=36.87^\circ\), \(143.13^\circ\), \(396.87^\circ\), \(503.13^\circ\). Halving: \(x=18.4^\circ\), \(71.6^\circ\), \(198.4^\circ\), \(251.6^\circ\). There are four solutions because the argument covers two full turns. The rule is: double the range, solve, then divide.

Worked example 16: using a compound angle in an equation

Solve \(\cos(x+30^\circ)=\sin x\) for \(0^\circ\le x<360^\circ\).

Expand: \(\cos x\cos30^\circ-\sin x\sin30^\circ=\sin x\), so \(\dfrac{\sqrt3}2\cos x=\tfrac32\sin x\), giving \(\tan x=\dfrac{\sqrt3/2}{3/2}=\dfrac1{\sqrt3}\). So \(x=30^\circ\) or \(210^\circ\). Check \(x=30^\circ\): \(\cos60^\circ=0.5=\sin30^\circ\).

Worked example 17: rewriting and then solving

Solve \(3\cos2x-5\cos x+2=0\) for \(0^\circ\le x<360^\circ\).

Use \(\cos2x=2\cos^2x-1\): \(6\cos^2x-3-5\cos x+2=0\), so \(6\cos^2x-5\cos x-1=0\). This factorises as \((6\cos x+1)(\cos x-1)=0\). \(\cos x=1\) gives \(x=0^\circ\). \(\cos x=-\tfrac16\) gives \(x=99.6^\circ\) or \(260.4^\circ\). Solutions: \(0^\circ\), \(99.6^\circ\), \(260.4^\circ\).

Worked example 18: R form with a negative coefficient

Write \(2\sin x-5\cos x\) in the form \(R\sin(x-\alpha)\) with \(\alpha\) acute, and find its least value.

\(R\sin(x-\alpha)=R\sin x\cos\alpha-R\cos x\sin\alpha\), so \(R\cos\alpha=2\) and \(R\sin\alpha=5\). Then \(R=\sqrt{29}=5.385\) and \(\tan\alpha=\dfrac52\), so \(\alpha=68.20^\circ\). The expression is \(\sqrt{29}\sin(x-68.20^\circ)\), whose least value is \(-\sqrt{29}=-5.39\). It occurs when \(x-68.20^\circ=270^\circ\), that is \(x=338.2^\circ\).

Worked example 19: tan of a sum in exact form

Given that \(\tan A=2\) and \(\tan B=3\), find \(\tan(A+B)\) and explain the sign of the answer.

\(\tan(A+B)=\dfrac{2+3}{1-2\times3}=\dfrac5{-5}=-1\). Both angles are acute, since their tangents are positive, and \(A=63.4^\circ\), \(B=71.6^\circ\). The sum is \(135^\circ\), an obtuse angle, where tangent is negative. Always check that the sign of the result agrees with the size of the angle.