TMUA: Trig graphs, identities and equations
Trig graphs, the two identities and solving trig equations.
- Section 1 Part 1 · MM4 Trigonometry · spec MM4.4-MM4.6
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- Graphs, symmetry and period
- tan = sin/cos and sin² + cos² = 1
- Counting solutions in an interval; quadratics in sin or cos
Key ideas
- \(\sin\) and \(\cos\) have period \(2\pi\), \(\tan\) has period \(\pi\). Use symmetry: \(\sin(\pi-x)=\sin x\), \(\cos(2\pi-x)=\cos x\).
- \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta+\cos^2\theta=1\).
- Turn a mixed equation into a quadratic in one function, e.g. \(2\sin^2x=3\cos x+3\Rightarrow 2\cos^2x+3\cos x+1=0\).
- Count solutions on a sketch: \(\cos x=c\) with \(-1
Common mistakes
- Dividing by \(\cos x\) or \(\sin x\) can lose solutions where it is zero.
- Watch the interval: \(0\le x\le4\pi\) is two full periods.
Exam tip
Sketch the graph over the whole interval and count crossings; it is quicker than listing angles.
Worked example
Worked example
How many solutions does the equation \(2\sin^2x=3\cos x+3\) have in the interval \(0\le x\le 4\pi\)?
- \(8\)
- \(3\)
- \(6\)
- \(4\)
- \(5\)
Answer: C: \(6\)
Use \(\sin^2x=1-\cos^2x\): \(2\cos^2x+3\cos x+1=0\), so \((2\cos x+1)(\cos x+1)=0\). \(\cos x=-\tfrac12\) has 2 solutions in each period (4 in \([0,4\pi]\)); \(\cos x=-1\) gives \(x=\pi,3\pi\). Total 6.
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
What is the greatest value of \(4\sin x-\cos^2x\) as \(x\) varies over the real numbers?
- \(5\)
- \(3\)
- \(1\)
- \(\frac{17}{4}\)
- \(4\)
Show the answer and solution
Answer: E: \(4\)
With \(s=\sin x\in[-1,1]\): \(4s-(1-s^2)=s^2+4s-1=(s+2)^2-5\). This increases for \(s\ge-2\), so on \([-1,1]\) the maximum is at \(s=1\): \(9-5=4\).
Question 2
How many solutions does the equation \(2\cos^2x+\cos x=0\) have in the interval \(0\le x\le 2\pi\)?
- \(5\)
- \(3\)
- \(4\)
- \(6\)
- \(8\)
Show the answer and solution
Answer: C: \(4\)
The equation factorises as \((2\cos x+1)(\cos x)=0\), so \(\cos x=- \frac{1}{2}\) or \(\cos x=0\). Counting the solutions of each on \([0,2\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 4 in total.
Question 3
How many solutions does the equation \(2\cos^2x+\cos x=0\) have in the interval \(0\le x\le 4\pi\)?
- \(7\)
- \(10\)
- \(9\)
- \(8\)
- \(16\)
Show the answer and solution
Answer: D: \(8\)
The equation factorises as \((2\cos x+1)(\cos x)=0\), so \(\cos x=- \frac{1}{2}\) or \(\cos x=0\). Counting the solutions of each on \([0,4\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 8 in total.
Question 4
How many solutions does the equation \(2\cos^2x-3\cos x+1=0\) have in the interval \(0\le x\le 3\pi\)?
- \(4\)
- \(5\)
- \(6\)
- \(10\)
- \(7\)
Show the answer and solution
Answer: B: \(5\)
The equation factorises as \((2\cos x-1)(\cos x-1)=0\), so \(\cos x=\frac{1}{2}\) or \(\cos x=1\). Counting the solutions of each on \([0,3\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 5 in total.
Question 5
How many solutions does the equation \(2\cos^2x+5\cos x+2=0\) have in the interval \(0\le x\le 4\pi\)?
- \(4\)
- \(8\)
- \(5\)
- \(3\)
- \(6\)
Show the answer and solution
Answer: A: \(4\)
The equation factorises as \((2\cos x+1)(\cos x+2)=0\), so \(\cos x=- \frac{1}{2}\) or \(\cos x=-2\). Counting the solutions of each on \([0,4\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 4 in total.
Question 6
How many solutions does the equation \(2\cos^2x-\cos x=0\) have in the interval \(0\le x\le 3\pi\)?
- \(7\)
- \(12\)
- \(6\)
- \(8\)
- \(5\)
Show the answer and solution
Answer: C: \(6\)
The equation factorises as \((2\cos x-1)(\cos x)=0\), so \(\cos x=\frac{1}{2}\) or \(\cos x=0\). Counting the solutions of each on \([0,3\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 6 in total.
Question 7
How many solutions does the equation \(2\cos^2x+3\cos x+1=0\) have in the interval \(0\le x\le 3\pi\)?
- \(6\)
- \(7\)
- \(5\)
- \(10\)
- \(4\)
Show the answer and solution
Answer: C: \(5\)
The equation factorises as \((2\cos x+1)(\cos x+1)=0\), so \(\cos x=- \frac{1}{2}\) or \(\cos x=-1\). Counting the solutions of each on \([0,3\pi]\) (a value of \(\cos\) strictly between \(-1\) and \(1\) is taken twice per period \(2\pi\); \(\pm1\) once; values outside \([-1,1]\) never) gives 5 in total.
Keep going
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- Next topic: Exponential graphs and laws of logarithms
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