What is 5.5: Trig identities?

The quotient identity is tanθ=sinθ/cosθ, and the three Pythagorean identities are sin^2θ+cos^2θ=1, ^2θ=1+tan^2θ and ^2θ=1+^2θ.

How many marks are 5.5 questions worth?

5.5 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 5.5 appeared in?

We have not yet indexed a verified past-paper appearance for 5.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.5

Understand and use tan θ = sin θ / cos θ; understand and use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.

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Trig identities

Worked answers and methods for 5.5 on Edexcel A-level Maths 9MA0.

Explanation

The quotient identity is $\tan\theta=\sin\theta/\cos\theta$ , and the three Pythagorean identities are $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta=1+\tan^2\theta$ and $\cosec^2\theta=1+\cot^2\theta$ .
Choose an identity containing the known and required functions, rearrange it, and use the stated quadrant to select the correct sign after taking a square root.
If $\tan\theta=-7/24$ in quadrant II, a reference triangle has side magnitudes $7$ , $24$ and $25$ , so $\sin\theta=7/25$ and $\cos\theta=-24/25$ .
From $\sin^2\theta+\cos^2\theta=1$ one obtains two possible signs; ignoring the quadrant or silently choosing the positive root can change every reciprocal ratio that follows.

Worked example

Given that $\tan\theta=-7/24$ and $\pi/2<\theta<\pi$ , determine $\sin\theta$ , $\cos\theta$ and $\cosec\theta$ .

1.Use $\sec^2\theta=1+\tan^2\theta=1+49/576=625/576$ .
2.In quadrant II cosine and secant are negative, so $\sec\theta=-25/24$ and $\cos\theta=-24/25$ .
3.Then $\sin\theta=\tan\theta\cos\theta=(-7/24)(-24/25)=7/25$ , giving $\cosec\theta=25/7$ .

Answer: $\sin\theta=7/25$ $\cos\theta=-24/25$ $\cosec\theta=25/7$

Common mistakes

Don't treat $\cosec\theta$ as $\sin\theta$ rather than its reciprocal.
Don't use $\tan\theta=\cos\theta/\sin\theta$ instead of $\sin\theta/\cos\theta$ .

Exam tip

Use the interval to fix the signs of sine and cosine before taking reciprocals.

Worked practice

Tier 1 · Easy

An acute angle

\theta

satisfies

\sin\theta=5/13

. Find

\tan\theta

and

\sec\theta

exactly.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\tan\theta=5/12$ $\sec\theta=13/12$	3
Notes
Because $\theta$ is acute, $\cos\theta$ is positive. From $\cos^2\theta=1-25/169=144/169$ , $\cos\theta=12/13$ . Hence $\tan\theta=(5/13)/(12/13)=5/12$ and $\sec\theta=1/(12/13)=13/12$ .

(3 marks)

Tier 2 · Standard

Prove that

\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}=2\sec^2x

for values of

x

for which both sides are defined.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Both sides simplify to $\dfrac{2}{\cos^2x}$ .	4
Notes
Combine the fractions on the left: $\dfrac{(1+\sin x)+(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{2}{1-\sin^2x}$ . Using $1-\sin^2x=\cos^2x$ , this is $2/\cos^2x=2\sec^2x$ . The manipulation is valid wherever the original expressions are defined.

(4 marks)

Tier 3 · Hard

An angle

\theta

lies in the first quadrant and satisfies

\sec\theta+\tan\theta=5

. Find

\sin\theta

exactly.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\sin\theta=12/13$	5
Notes
Since $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$ , it follows that $\sec\theta-\tan\theta=1/5$ . Adding the two equations gives $2\sec\theta=26/5$ , so $\sec\theta=13/5$ and $\tan\theta=12/5$ . Thus $\cos\theta=5/13$ and $\sin\theta=\tan\theta\cos\theta=(12/5)(5/13)=12/13$ .

(5 marks)

Tier 1 · Easy

Given that

\cot\theta=-3/4

and

3\pi/2<\theta<2\pi

, find

\cosec\theta

exactly.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\cosec\theta=-\dfrac54$	2
Notes
Using a $3$ - $4$ - $5$ reference triangle, $\|\sin\theta\|=4/5$ . The angle is in quadrant IV, where sine is negative, so $\sin\theta=-4/5$ and $\cosec\theta=-5/4$ .

(2 marks)

Tier 2 · Standard

Simplify

\dfrac{\sec x-\cos x}{\tan x}

, for values of

x

for which the original expression is defined.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\sin x$	3
Notes
$\sec x-\cos x=1/\cos x-\cos x=(1-\cos^2x)/\cos x=\sin^2x/\cos x$ . Dividing this by $\tan x=\sin x/\cos x$ gives $\sin x$ wherever the original expression is defined.

(3 marks)

Tier 3 · Hard

Given that

\sin\theta+\cos\theta=7/5

and

0<\theta<\pi/4

, find

\tan\theta

exactly.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\tan\theta=\dfrac34$	5
Notes
Squaring the given relation gives $1+2\sin\theta\cos\theta=49/25$ , so $\sin\theta\cos\theta=12/25$ . Therefore $(\cos\theta-\sin\theta)^2=1-2\sin\theta\cos\theta=1/25$ . In the stated interval $\cos\theta>\sin\theta$ , so $\cos\theta-\sin\theta=1/5$ . Solving this with $\sin\theta+\cos\theta=7/5$ gives $\sin\theta=3/5$ and $\cos\theta=4/5$ . Hence $\tan\theta=3/4$ .

(5 marks)

Tier 2 · Standard

Given that

\tan\theta=-2

and

3\pi/2<\theta<2\pi

, find the exact value of

\dfrac{3\sin\theta+\cos\theta}{\sin\theta-2\cos\theta}

without finding

\theta

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$5/4$	3
Notes
In the stated quadrant $\cos\theta\neq0$ , so divide the numerator and denominator by $\cos\theta$ . The expression becomes $\dfrac{3\tan\theta+1}{\tan\theta-2}$ . Substituting $\tan\theta=-2$ gives $(-6+1)/(-2-2)=5/4$ .

(3 marks)

Tier 3 · Hard

An acute angle

\theta

satisfies

2\sec\theta+\tan\theta=4

. Find

\sin\theta

exactly.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\sin\theta=\dfrac{4\sqrt{13}-2}{17}$	5
Notes
Let $s=\sec\theta$ and $t=\tan\theta$ . Then $t=4-2s$ and $s^2=1+t^2$ , so $s^2=1+(4-2s)^2$ . Hence $3s^2-16s+17=0$ , giving $s=(8\pm\sqrt{13})/3$ . Since $\theta$ is acute, $t>0$ ; the plus sign would give $t=4-2s<0$ , so $s=(8-\sqrt{13})/3$ . Thus $\cos\theta=1/s=(8+\sqrt{13})/17$ , and $\sin\theta=\tan\theta\cos\theta=(4-2s)/s=4\cos\theta-2=(4\sqrt{13}-2)/17$ .

(5 marks)

Tier 3 · Hard

An angle

\theta

satisfies

\pi/2<\theta<\pi

and

\dfrac{2\sin\theta+\cos\theta}{\sin\theta-2\cos\theta}=\dfrac{2}{11}

. Find

\cosec\theta-\cot\theta

exactly.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\cosec\theta-\cot\theta=3$	5
Notes
In the stated interval $\cos\theta\neq0$ , so divide the numerator and denominator by $\cos\theta$ . With $t=\tan\theta$ , the equation becomes $(2t+1)/(t-2)=2/11$ . Cross-multiplying gives $22t+11=2t-4$ , hence $t=-3/4$ . Since $\theta$ is in quadrant II, a $3$ - $4$ - $5$ reference triangle gives $\sin\theta=3/5$ and $\cos\theta=-4/5$ . Therefore $\cosec\theta-\cot\theta=5/3-(-4/3)=3$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Prove that

\sec^2x+\cosec^2x\geq4

wherever the expression is defined. Find all values of

x

for which equality holds in the interval

0\leq x<2\pi

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\sec^2x+\cosec^2x\geq4$ $x=\pi/4,\ 3\pi/4,\ 5\pi/4,\ 7\pi/4$	5
Notes
Where the expression is defined, $\sin x\cos x\neq0$ and $\sec^2x+\cosec^2x=(\sin^2x+\cos^2x)/(\sin^2x\cos^2x)=1/(\sin^2x\cos^2x)$ . Also $(\sin^2x-\cos^2x)^2\geq0$ , so $(\sin^2x+\cos^2x)^2\geq4\sin^2x\cos^2x$ . Thus $\sin^2x\cos^2x\leq1/4$ , proving the inequality. Equality requires $\sin^2x=\cos^2x$ , which gives the four stated values in the interval.

(5 marks)

Verified exam appearances

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