Skip to content
Oscar Song Tutoring
5.5

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

Draft — not yet indexed

Trig identities

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

Explanation

  • The quotient identity is tan⁡θ=sin⁡θ/cos⁡θ\tan\theta=\sin\theta/\cos\theta, and the three Pythagorean identities are sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta and cosec⁡2θ=1+cot⁡2θ\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⁡θ=−7/24\tan\theta=-7/24 in quadrant II, a reference triangle has side magnitudes 77, 2424 and 2525, so sin⁡θ=7/25\sin\theta=7/25 and cos⁡θ=−24/25\cos\theta=-24/25.
  • From sin⁡2θ+cos⁡2θ=1\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⁡θ=−7/24\tan\theta=-7/24 and π/2<θ<π\pi/2<\theta<\pi, determine sin⁡θ\sin\theta, cos⁡θ\cos\theta and cosec⁡θ\cosec\theta.

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

Answer: sin⁡θ=7/25\sin\theta=7/25 cos⁡θ=−24/25\cos\theta=-24/25 cosec⁡θ=25/7\cosec\theta=25/7

Common mistakes

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

Exam tip

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

Worked practice

Q1
Tier 1 · Easy

1.

An acute angle θ\theta satisfies sin⁡θ=5/13\sin\theta=5/13. Find tan⁡θ\tan\theta and sec⁡θ\sec\theta exactly.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • tan⁡θ=5/12\tan\theta=5/12
  • sec⁡θ=13/12\sec\theta=13/12
3
Notes
Because θ\theta is acute, cos⁡θ\cos\theta is positive. From cos⁡2θ=1−25/169=144/169\cos^2\theta=1-25/169=144/169, cos⁡θ=12/13\cos\theta=12/13. Hence tan⁡θ=(5/13)/(12/13)=5/12\tan\theta=(5/13)/(12/13)=5/12 and sec⁡θ=1/(12/13)=13/12\sec\theta=1/(12/13)=13/12.

(3 marks)

Q2
Tier 2 · Standard

2.

Prove that 11−sin⁡x+11+sin⁡x=2sec⁡2x\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}=2\sec^2x for values of xx for which both sides are defined.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • Both sides simplify to 2cos⁡2x\dfrac{2}{\cos^2x}.
4
Notes
Combine the fractions on the left: (1+sin⁡x)+(1−sin⁡x)(1−sin⁡x)(1+sin⁡x)=21−sin⁡2x\dfrac{(1+\sin x)+(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{2}{1-\sin^2x}. Using 1−sin⁡2x=cos⁡2x1-\sin^2x=\cos^2x, this is 2/cos⁡2x=2sec⁡2x2/\cos^2x=2\sec^2x. The manipulation is valid wherever the original expressions are defined.

(4 marks)

Q3
Tier 3 · Hard

3.

An angle θ\theta lies in the first quadrant and satisfies sec⁡θ+tan⁡θ=5\sec\theta+\tan\theta=5. Find sin⁡θ\sin\theta exactly.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • sin⁡θ=12/13\sin\theta=12/13
5
Notes
Since (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=sec⁡2θ−tan⁡2θ=1(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1, it follows that sec⁡θ−tan⁡θ=1/5\sec\theta-\tan\theta=1/5. Adding the two equations gives 2sec⁡θ=26/52\sec\theta=26/5, so sec⁡θ=13/5\sec\theta=13/5 and tan⁡θ=12/5\tan\theta=12/5. Thus cos⁡θ=5/13\cos\theta=5/13 and sin⁡θ=tan⁡θcos⁡θ=(12/5)(5/13)=12/13\sin\theta=\tan\theta\cos\theta=(12/5)(5/13)=12/13.

(5 marks)

Q4
Tier 1 · Easy

4.

Given that cot⁡θ=−3/4\cot\theta=-3/4 and 3π/2<θ<2π3\pi/2<\theta<2\pi, find cosec⁡θ\cosec\theta exactly.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • cosec⁡θ=−54\cosec\theta=-\dfrac54
2
Notes
Using a 33-44-55 reference triangle, ∣sin⁡θ∣=4/5|\sin\theta|=4/5. The angle is in quadrant IV, where sine is negative, so sin⁡θ=−4/5\sin\theta=-4/5 and cosec⁡θ=−5/4\cosec\theta=-5/4.

(2 marks)

Q5
Tier 2 · Standard

5.

Simplify sec⁡x−cos⁡xtan⁡x\dfrac{\sec x-\cos x}{\tan x}, for values of xx for which the original expression is defined.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • sin⁡x\sin x
3
Notes
sec⁡x−cos⁡x=1/cos⁡x−cos⁡x=(1−cos⁡2x)/cos⁡x=sin⁡2x/cos⁡x\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\tan x=\sin x/\cos x gives sin⁡x\sin x wherever the original expression is defined.

(3 marks)

Q6
Tier 3 · Hard

6.

Given that sin⁡θ+cos⁡θ=7/5\sin\theta+\cos\theta=7/5 and 0<θ<π/40<\theta<\pi/4, find tan⁡θ\tan\theta exactly.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • tan⁡θ=34\tan\theta=\dfrac34
5
Notes
Squaring the given relation gives 1+2sin⁡θcos⁡θ=49/251+2\sin\theta\cos\theta=49/25, so sin⁡θcos⁡θ=12/25\sin\theta\cos\theta=12/25. Therefore (cos⁡θ−sin⁡θ)2=1−2sin⁡θcos⁡θ=1/25(\cos\theta-\sin\theta)^2=1-2\sin\theta\cos\theta=1/25. In the stated interval cos⁡θ>sin⁡θ\cos\theta>\sin\theta, so cos⁡θ−sin⁡θ=1/5\cos\theta-\sin\theta=1/5. Solving this with sin⁡θ+cos⁡θ=7/5\sin\theta+\cos\theta=7/5 gives sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=4/5\cos\theta=4/5. Hence tan⁡θ=3/4\tan\theta=3/4.

(5 marks)

Q7
Tier 2 · Standard

7.

Given that tan⁡θ=−2\tan\theta=-2 and 3π/2<θ<2π3\pi/2<\theta<2\pi, find the exact value of 3sin⁡θ+cos⁡θsin⁡θ−2cos⁡θ\dfrac{3\sin\theta+\cos\theta}{\sin\theta-2\cos\theta} without finding θ\theta.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • 5/45/4
3
Notes
In the stated quadrant cos⁡θ≠0\cos\theta\neq0, so divide the numerator and denominator by cos⁡θ\cos\theta. The expression becomes 3tan⁡θ+1tan⁡θ−2\dfrac{3\tan\theta+1}{\tan\theta-2}. Substituting tan⁡θ=−2\tan\theta=-2 gives (−6+1)/(−2−2)=5/4(-6+1)/(-2-2)=5/4.

(3 marks)

Q8
Tier 3 · Hard

8.

An acute angle θ\theta satisfies 2sec⁡θ+tan⁡θ=42\sec\theta+\tan\theta=4. Find sin⁡θ\sin\theta exactly.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • sin⁡θ=413−217\sin\theta=\dfrac{4\sqrt{13}-2}{17}
5
Notes
Let s=sec⁡θs=\sec\theta and t=tan⁡θt=\tan\theta. Then t=4−2st=4-2s and s2=1+t2s^2=1+t^2, so s2=1+(4−2s)2s^2=1+(4-2s)^2. Hence 3s2−16s+17=03s^2-16s+17=0, giving s=(8±13)/3s=(8\pm\sqrt{13})/3. Since θ\theta is acute, t>0t>0; the plus sign would give t=4−2s<0t=4-2s<0, so s=(8−13)/3s=(8-\sqrt{13})/3. Thus cos⁡θ=1/s=(8+13)/17\cos\theta=1/s=(8+\sqrt{13})/17, and sin⁡θ=tan⁡θcos⁡θ=(4−2s)/s=4cos⁡θ−2=(413−2)/17\sin\theta=\tan\theta\cos\theta=(4-2s)/s=4\cos\theta-2=(4\sqrt{13}-2)/17.

(5 marks)

Q9
Tier 3 · Hard

9.

An angle θ\theta satisfies π/2<θ<π\pi/2<\theta<\pi and 2sin⁡θ+cos⁡θsin⁡θ−2cos⁡θ=211\dfrac{2\sin\theta+\cos\theta}{\sin\theta-2\cos\theta}=\dfrac{2}{11}. Find cosec⁡θ−cot⁡θ\cosec\theta-\cot\theta exactly.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • cosec⁡θ−cot⁡θ=3\cosec\theta-\cot\theta=3
5
Notes
In the stated interval cos⁡θ≠0\cos\theta\neq0, so divide the numerator and denominator by cos⁡θ\cos\theta. With t=tan⁡θt=\tan\theta, the equation becomes (2t+1)/(t−2)=2/11(2t+1)/(t-2)=2/11. Cross-multiplying gives 22t+11=2t−422t+11=2t-4, hence t=−3/4t=-3/4. Since θ\theta is in quadrant II, a 33-44-55 reference triangle gives sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=−4/5\cos\theta=-4/5. Therefore cosec⁡θ−cot⁡θ=5/3−(−4/3)=3\cosec\theta-\cot\theta=5/3-(-4/3)=3.

(5 marks)

Q10
Tier 3 · Hard

10.

Prove that sec⁡2x+cosec⁡2x≥4\sec^2x+\cosec^2x\geq4 wherever the expression is defined. Find all values of xx for which equality holds in the interval 0≤x<2π0\leq x<2\pi.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • sec⁡2x+cosec⁡2x≥4\sec^2x+\cosec^2x\geq4
  • x=π/4, 3π/4, 5π/4, 7π/4x=\pi/4,\ 3\pi/4,\ 5\pi/4,\ 7\pi/4
5
Notes
Where the expression is defined, sin⁡xcos⁡x≠0\sin x\cos x\neq0 and sec⁡2x+cosec⁡2x=(sin⁡2x+cos⁡2x)/(sin⁡2xcos⁡2x)=1/(sin⁡2xcos⁡2x)\sec^2x+\cosec^2x=(\sin^2x+\cos^2x)/(\sin^2x\cos^2x)=1/(\sin^2x\cos^2x). Also (sin⁡2x−cos⁡2x)2≥0(\sin^2x-\cos^2x)^2\geq0, so (sin⁡2x+cos⁡2x)2≥4sin⁡2xcos⁡2x(\sin^2x+\cos^2x)^2\geq4\sin^2x\cos^2x. Thus sin⁡2xcos⁡2x≤1/4\sin^2x\cos^2x\leq1/4, proving the inequality. Equality requires sin⁡2x=cos⁡2x\sin^2x=\cos^2x, which gives the four stated values in the interval.

(5 marks)

Q11
Tier 2 · Standard

11.

(a) Prove that 1−cos⁡2θcos⁡θ≡sin⁡θtan⁡θ\dfrac{1-\cos^2\theta}{\cos\theta}\equiv\sin\theta\tan\theta, for cos⁡θ≠0\cos\theta\neq0. (b) Given that θ\theta is obtuse and sin⁡θ=35\sin\theta=\dfrac35, find the exact value of tan⁡θ\tan\theta.

(5)

(Total for Question 11 is 5 marks)

Mark scheme

Mark scheme for question 11
QuestionSchemeMarks
11
  • (a) 1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=sin⁡θ×sin⁡θcos⁡θ=sin⁡θtan⁡θ\dfrac{1-\cos^2\theta}{\cos\theta}=\dfrac{\sin^2\theta}{\cos\theta}=\sin\theta\times\dfrac{\sin\theta}{\cos\theta}=\sin\theta\tan\theta
  • (b) tan⁡θ=−34\tan\theta=-\dfrac34
5
Notes
(a) Since sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, the numerator 1−cos⁡2θ1-\cos^2\theta equals sin⁡2θ\sin^2\theta. Then sin⁡2θcos⁡θ=sin⁡θ×sin⁡θcos⁡θ=sin⁡θtan⁡θ\dfrac{\sin^2\theta}{\cos\theta}=\sin\theta\times\dfrac{\sin\theta}{\cos\theta}=\sin\theta\tan\theta, using tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}. (b) cos⁡2θ=1−(35)2=1625\cos^2\theta=1-\left(\tfrac35\right)^2=\tfrac{16}{25}, so cos⁡θ=±45\cos\theta=\pm\tfrac45. An obtuse angle has a negative cosine, so cos⁡θ=−45\cos\theta=-\tfrac45 and tan⁡θ=3/5−4/5=−34\tan\theta=\dfrac{3/5}{-4/5}=-\tfrac34.

(5 marks)

Verified exam appearances

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

Other points in 5 Trigonometry

Want help turning this into marks?

Bring 5.5 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call