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The identities \(\sin^2\theta + \cos^2\theta \equiv 1\) and \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) let you find exact values and solve equations. They appear in almost every trigonometry question on Paper 1.
\(\sin^2\theta\) means \((\sin\theta)^2\). The symbol ≡ means the statement is true for every value of θ for which it is defined.
Exact values
diagram
Given one ratio, use an identity to find another, then choose the sign from the quadrant (CAST).
Example: if \(\sin\theta = -\dfrac{5}{13}\) and \(180^\circ \lt \theta \lt 270^\circ\), then \(\cos^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}\). cos is negative in this quadrant, so \(\cos\theta = -\dfrac{12}{13}\)
You can also draw a right-angled triangle for the acute angle (here with sides 5, 12 and 13), then add the signs.
For the acute angle α: sin α = 5/13, cos α = 12/13, tan α = 5/12. Then add the signs for the quadrant.
Solving equations
Aim for an equation in one trig function: replace \(\cos^2 x\) with \(1 - \sin^2 x\) if the equation also has sin x, or \(\tan^2 x\) with \(\sec^2 x - 1\) if it also has sec x.
Rearrange into a quadratic equal to 0, factorise (or use the formula), then solve each factor in the given interval.
Reject impossible values: \(\sin x = 2\) and \(\cos x = -3\) have no solutions, and neither has \(\sec x = \dfrac{1}{2}\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, ft = follow through: you still earn the mark if you use your own earlier wrong answer correctly.
Find exact values from one given ratio
4 marksGrade C
Use \(\sin^2\theta + \cos^2\theta = 1\) to find the square of the missing ratio.
Choose the sign using the quadrant.
Use reciprocals and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) for the other ratios.
Example. Given that \(\sin\theta = -\dfrac{5}{13}\) and \(180^\circ \lt \theta \lt 270^\circ\), find the exact value of (a) \(\cos\theta\) (b) \(\sec\theta\) (c) \(\tan\theta\)
Solve a quadratic equation using \(\sin^2x+\cos^2x=1\)
5 marksGrade B
Replace \(\cos^2 x\) with \(1 - \sin^2 x\) so that only sin x is left.
Rearrange into a quadratic equal to zero and factorise.
Solve each factor in the interval, rejecting any impossible value.
Example. Solve, for \(0 \le x \lt 360^\circ\), the equation \(2\cos^2 x = 1 + \sin x\)
Show the model answerHide the model answer
\(2(1 - \sin^2 x) = 1 + \sin x\)
M1
\(2\sin^2 x + \sin x - 1 = 0\)
A1
\((2\sin x - 1)(\sin x + 1) = 0\), so \(\sin x = \dfrac{1}{2}\) or \(\sin x = -1\)
M1 A1
x = 30°, 150°, 270°
A1
Don’t lose marks
Writing \(\cos^2 x = 1 - \sin x\) (the square is lost).
Giving a positive value for cos θ when θ is in the second or third quadrant.
Not rejecting impossible values such as \(\sin x = -2\)
Missing solutions at the ends of the interval, such as θ = 0°.
More tips
Memory tricks
Learn \(\sin^2\theta + \cos^2\theta = 1\) well: divide it by \(\cos^2\theta\) to get the tan and sec identity, or by \(\sin^2\theta\) to get the cot and cosec one.
The 3, 4, 5 and 5, 12, 13 triangles give exact values quickly.
Exam technique
Show the line where you substitute the identity; it usually earns the first method mark.
Give angles to 1 d.p. in degrees (or 3 s.f. in radians) unless you are told otherwise.
When the question says 'exact', leave answers as fractions or surds.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade D
Use \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) to simplifye.g. \(\sin\theta = 2\cos\theta\) becomes \(\tan\theta = 2\)
Grade C
Find exact values using \(\sin^2\theta+\cos^2\theta=1\)Given one ratio and the quadrant, find the other ratios with the correct signs.
Grade B
Solve quadratic equations using an identityReplace \(\cos^2 x\) by \(1 - \sin^2 x\) (or similar) to get a quadratic in one function.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Show that the equation \[2\cos^2 x + 5\sin x = -1\] can be written in the form \[2\sin^2 x - 5\sin x - 3 = 0.\][2]
(b) Hence solve, for \(0 \le x \lt 360^\circ\), the equation \[2\cos^2 x + 5\sin x = -1.\][4]
Show the answer and mark scheme
(a)Answer: \(2\sin^2 x - 5\sin x - 3 = 0\) (shown)
M1 for using \(\cos^2 x = 1 - \sin^2 x\)
A1* for \(2\sin^2 x - 5\sin x - 3 = 0\) reached with no errors
Worked solution: \(2(1 - \sin^2 x) + 5\sin x + 1 = 0\) \(\Rightarrow -2\sin^2 x + 5\sin x + 3 = 0 \Rightarrow 2\sin^2 x - 5\sin x - 3 = 0\)
(b)Answer: \(x = 210^\circ, \ 330^\circ\) °
M1 for solving the quadratic in \(\sin x\), e.g. \((2\sin x + 1)(\sin x - 3) = 0\)
A1 for \(\sin x = -\frac{1}{2}\), with \(\sin x = 3\) rejected or ignored
dM1 for a correct method to find a second solution in the interval (using the symmetry of the graph of \(y = \sin x\))
A1 for \(210^\circ, \ 330^\circ\) and no other values in the interval
Worked solution: \((2\sin x + 1)(\sin x - 3) = 0\), so \(\sin x = -\frac{1}{2}\) or \(\sin x = 3\). Since \(-1 \le \sin x \le 1\), \(\sin x = 3\) has no solutions. \(\sin x = -\frac{1}{2}\): \(\arcsin\left(-\frac{1}{2}\right) = -30^\circ\), giving \(x = 210^\circ, \ 330^\circ\) So \(x = 210^\circ, \ 330^\circ\).
Question 2Medium6 marks
(a) Show that the equation \[8\sin^2 x - 14\cos x - 13 = 0\] can be written in the form \[8\cos^2 x + 14\cos x + 5 = 0.\][2]
(b) Hence solve, for \(-180^\circ \le x \lt 180^\circ\), the equation \[8\sin^2 x - 14\cos x - 13 = 0.\][4]
Show the answer and mark scheme
(a)Answer: \(8\cos^2 x + 14\cos x + 5 = 0\) (shown)
M1 for using \(\sin^2 x = 1 - \cos^2 x\)
A1* for \(8\cos^2 x + 14\cos x + 5 = 0\) reached with no errors
Worked solution: \(8(1 - \cos^2 x) - 14\cos x - 13 = 0\) \(\Rightarrow -8\cos^2 x - 14\cos x - 5 = 0 \Rightarrow 8\cos^2 x + 14\cos x + 5 = 0\)
(b)Answer: \(x = -120^\circ, \ 120^\circ\) °
M1 for solving the quadratic in \(\cos x\), e.g. \((2\cos x + 1)(4\cos x + 5) = 0\)
A1 for \(\cos x = -\frac{1}{2}\), with \(\cos x = -\frac{5}{4}\) rejected or ignored
dM1 for a correct method to find a second solution in the interval (using the symmetry of the graph of \(y = \cos x\))
A1 for \(-120^\circ, \ 120^\circ\) and no other values in the interval
Worked solution: \((2\cos x + 1)(4\cos x + 5) = 0\), so \(\cos x = -\frac{1}{2}\) or \(\cos x = -\frac{5}{4}\). Since \(-1 \le \cos x \le 1\), \(\cos x = -\frac{5}{4}\) has no solutions. \(\cos x = -\frac{1}{2}\): \(\arccos\left(-\frac{1}{2}\right) = 120^\circ\), giving \(x = -120^\circ, \ 120^\circ\) So \(x = -120^\circ, \ 120^\circ\).