Skip to content
Oscar Song Tutoring
5.8

Construct proofs involving trigonometric functions and identities.

Draft — not yet indexed

Trig proofs

Worked answers and methods for 5.8 on Edexcel A-level Maths 9MA0.

Explanation

  • A trigonometric identity is true for every value in its domain, so a proof transforms one side into the other using exact algebra and established identities rather than testing selected angles.
  • Usually begin with the more complicated side, replace secant, cosecant, cotangent or tangent by sine and cosine when helpful, and factor or take a common denominator before cancelling.
  • Multiplying numerator and denominator by a conjugate can expose 1−sin⁡2x=cos⁡2x1-\sin^2x=\cos^2x or 1−cos⁡2x=sin⁡2x1-\cos^2x=\sin^2x and complete the proof cleanly.
  • Never cancel terms across addition, and record domain restrictions: algebra such as division by sin⁡x\sin x is valid only where that denominator is non-zero.

Worked example

Prove that 1−cos⁡(2x)sin⁡(2x)=tan⁡x\dfrac{1-\cos(2x)}{\sin(2x)}=\tan x for values at which both sides are defined.

  1. 1.Apply the double-angle forms to the left-hand side: (1−cos⁡2x)/sin⁡2x=(2sin⁡2x)/(2sin⁡xcos⁡x)=sin⁡x/cos⁡x=tan⁡x(1-\cos2x)/\sin2x=(2\sin^2x)/(2\sin x\cos x)=\sin x/\cos x=\tan x, with cancellation only where the original expressions are defined.

Answer: Use 1−cos⁡(2x)=2sin⁡2x1-\cos(2x)=2\sin^2x and sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x.

Common mistakes

  • Don't cancel a term across addition while manipulating a trigonometric fraction.
  • Don't divide by sin⁡x\sin x or cos⁡x\cos x without recording where that factor is zero.

Exam tip

For “prove”, transform one side only with named identities until it exactly matches the other side.

Worked practice

Q1
Tier 1 · Easy

1.

Prove that sin⁡xcot⁡x=cos⁡x\sin x\cot x=\cos x wherever the left-hand side is defined.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • Replace cot⁡x\cot x by cos⁡x/sin⁡x\cos x/\sin x and cancel sin⁡x\sin x.
2
Notes
sin⁡xcot⁡x=sin⁡x(cos⁡x/sin⁡x)=cos⁡x\sin x\cot x=\sin x(\cos x/\sin x)=\cos x. The cancellation is valid wherever cot⁡x\cot x is defined, namely where sin⁡x≠0\sin x\neq0.

(2 marks)

Q2
Tier 2 · Standard

2.

Prove that cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • Expand both cosine factors and simplify using sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1.
4
Notes
Expand the left-hand side: (cos⁡Acos⁡B−sin⁡Asin⁡B)(cos⁡Acos⁡B+sin⁡Asin⁡B)=cos⁡2Acos⁡2B−sin⁡2Asin⁡2B(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)=\cos^2A\cos^2B-\sin^2A\sin^2B. Now use cos⁡2B=1−sin⁡2B\cos^2B=1-\sin^2B: this becomes cos⁡2A−sin⁡2B(cos⁡2A+sin⁡2A)=cos⁡2A−sin⁡2B\cos^2A-\sin^2B(\cos^2A+\sin^2A)=\cos^2A-\sin^2B.

(4 marks)

Q3
Tier 3 · Hard

3.

Prove that 11−sin⁡x−11+sin⁡x=2tan⁡xsec⁡x\dfrac{1}{1-\sin x}-\dfrac{1}{1+\sin x}=2\tan x\sec x wherever the expressions exist.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • The left-hand side simplifies to 2sin⁡x/cos⁡2x=2tan⁡xsec⁡x2\sin x/\cos^2x=2\tan x\sec x.
5
Notes
Combine the fractions: the numerator is (1+sin⁡x)−(1−sin⁡x)=2sin⁡x(1+\sin x)-(1-\sin x)=2\sin x and the denominator is (1−sin⁡x)(1+sin⁡x)=1−sin⁡2x=cos⁡2x(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x. Thus the left-hand side is 2sin⁡x/cos⁡2x=2(sin⁡x/cos⁡x)(1/cos⁡x)=2tan⁡xsec⁡x2\sin x/\cos^2x=2(\sin x/\cos x)(1/\cos x)=2\tan x\sec x.

(5 marks)

Q4
Tier 1 · Easy

4.

Prove that (1+cot⁡2x)sin⁡2x=1(1+\cot^2x)\sin^2x=1 wherever the left-hand side is defined.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • Use 1+cot⁡2x=cosec⁡2x1+\cot^2x=\cosec^2x.
2
Notes
By the Pythagorean identity, (1+cot⁡2x)sin⁡2x=cosec⁡2xsin⁡2x=(1/sin⁡2x)sin⁡2x=1(1+\cot^2x)\sin^2x=\cosec^2x\sin^2x=(1/\sin^2x)\sin^2x=1 wherever sin⁡x≠0\sin x\neq0.

(2 marks)

Q5
Tier 2 · Standard

5.

Prove that tan⁡xsec⁡x−1=1+cos⁡xsin⁡x\dfrac{\tan x}{\sec x-1}=\dfrac{1+\cos x}{\sin x} wherever both sides are defined.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • Multiply the left-hand side by (sec⁡x+1)/(sec⁡x+1)(\sec x+1)/(\sec x+1).
4
Notes
Multiply the left-hand side by the conjugate: tan⁡x(sec⁡x+1)sec⁡2x−1=tan⁡x(sec⁡x+1)tan⁡2x=sec⁡x+1tan⁡x\dfrac{\tan x(\sec x+1)}{\sec^2x-1}=\dfrac{\tan x(\sec x+1)}{\tan^2x}=\dfrac{\sec x+1}{\tan x}. Replacing secant and tangent by sine and cosine gives 1/cos⁡x+1sin⁡x/cos⁡x=1+cos⁡xsin⁡x\dfrac{1/\cos x+1}{\sin x/\cos x}=\dfrac{1+\cos x}{\sin x}, as required. The working multiplies by sec⁡x+1\sec x+1 and divides by tan⁡2x\tan^2x; both are nonzero wherever both sides are defined, since tan⁡x=0\tan x=0 or sec⁡x=−1\sec x=-1 would make a side undefined.

(4 marks)

Q6
Tier 3 · Hard

6.

Prove that (sin⁡x+cos⁡x)4−(sin⁡x−cos⁡x)4=4sin⁡2x(\sin x+\cos x)^4-(\sin x-\cos x)^4=4\sin2x.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • Factor the difference of fourth powers, then use sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 and sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x.
4
Notes
Let a=sin⁡x+cos⁡xa=\sin x+\cos x and b=sin⁡x−cos⁡xb=\sin x-\cos x. Then a4−b4=(a−b)(a+b)(a2+b2)a^4-b^4=(a-b)(a+b)(a^2+b^2). Here a−b=2cos⁡xa-b=2\cos x, a+b=2sin⁡xa+b=2\sin x, and a2+b2=2(sin⁡2x+cos⁡2x)=2a^2+b^2=2(\sin^2x+\cos^2x)=2. Thus the left-hand side is (2cos⁡x)(2sin⁡x)(2)=8sin⁡xcos⁡x=4sin⁡2x(2\cos x)(2\sin x)(2)=8\sin x\cos x=4\sin2x, as required.

(4 marks)

Q7
Tier 2 · Standard

7.

Prove that 1+sin⁡xcos⁡x+cos⁡x1+sin⁡x=2sec⁡x\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=2\sec x wherever both sides are defined.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • Combine the left-hand side over cos⁡x(1+sin⁡x)\cos x(1+\sin x) and use sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1.
4
Notes
The left-hand side is (1+sin⁡x)2+cos⁡2xcos⁡x(1+sin⁡x)\dfrac{(1+\sin x)^2+\cos^2x}{\cos x(1+\sin x)}. Its numerator simplifies to 1+2sin⁡x+sin⁡2x+cos⁡2x=2(1+sin⁡x)1+2\sin x+\sin^2x+\cos^2x=2(1+\sin x). Since the original expressions require cos⁡x≠0\cos x\neq0 and 1+sin⁡x≠01+\sin x\neq0, cancellation is valid, leaving 2/cos⁡x=2sec⁡x2/\cos x=2\sec x.

(4 marks)

Q8
Tier 3 · Hard

8.

Prove that sec⁡2x+cosec⁡2x=(tan⁡x+cot⁡x)2=4cosec⁡2(2x)\sec^2x+\cosec^2x=(\tan x+\cot x)^2=4\cosec^2(2x) wherever the expressions are defined.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • Each expression simplifies to 1/(sin⁡2xcos⁡2x)1/(\sin^2x\cos^2x).
5
Notes
Where the expressions are defined, sin⁡x≠0\sin x\neq0 and cos⁡x≠0\cos x\neq0. First, sec⁡2x+cosec⁡2x=1/cos⁡2x+1/sin⁡2x=(sin⁡2x+cos⁡2x)/(sin⁡2xcos⁡2x)=1/(sin⁡2xcos⁡2x)\sec^2x+\cosec^2x=1/\cos^2x+1/\sin^2x=(\sin^2x+\cos^2x)/(\sin^2x\cos^2x)=1/(\sin^2x\cos^2x). Also tan⁡x+cot⁡x=sin⁡x/cos⁡x+cos⁡x/sin⁡x=1/(sin⁡xcos⁡x)\tan x+\cot x=\sin x/\cos x+\cos x/\sin x=1/(\sin x\cos x), so its square is the same expression. Finally, 4cosec⁡2(2x)=4/sin⁡2(2x)=4/(4sin⁡2xcos⁡2x)4\cosec^2(2x)=4/\sin^2(2x)=4/(4\sin^2x\cos^2x), again giving the same result.

(5 marks)

Q9
Tier 3 · Hard

9.

Prove that sin⁡6x+cos⁡6x=1−34sin⁡2(2x)\sin^6x+\cos^6x=1-\dfrac34\sin^2(2x). Hence find the exact value of sin⁡6(π/8)+cos⁡6(π/8)\sin^6(\pi/8)+\cos^6(\pi/8).

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • sin⁡6x+cos⁡6x=1−34sin⁡2(2x)\sin^6x+\cos^6x=1-\dfrac34\sin^2(2x)
  • 5/85/8
5
Notes
Using a3+b3=(a+b)3−3ab(a+b)a^3+b^3=(a+b)^3-3ab(a+b) with a=sin⁡2xa=\sin^2x and b=cos⁡2xb=\cos^2x gives sin⁡6x+cos⁡6x=1−3sin⁡2xcos⁡2x\sin^6x+\cos^6x=1-3\sin^2x\cos^2x. Since sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x, this is 1−3sin⁡2(2x)/41-3\sin^2(2x)/4. At x=π/8x=\pi/8, sin⁡(2x)=sin⁡(π/4)=2/2\sin(2x)=\sin(\pi/4)=\sqrt2/2, so the value is 1−(3/4)(1/2)=5/81-(3/4)(1/2)=5/8.

(5 marks)

Q10
Tier 3 · Hard

10.

Prove that (sec⁡x+tan⁡x)2=1+sin⁡x1−sin⁡x(\sec x+\tan x)^2=\dfrac{1+\sin x}{1-\sin x} wherever both sides are defined. Hence solve sec⁡x+tan⁡x=3\sec x+\tan x=\sqrt3 for −π<x<π-\pi<x<\pi.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • (sec⁡x+tan⁡x)2=1+sin⁡x1−sin⁡x(\sec x+\tan x)^2=\dfrac{1+\sin x}{1-\sin x}
  • x=π/6x=\pi/6
6
Notes
Write sec⁡x+tan⁡x=(1+sin⁡x)/cos⁡x\sec x+\tan x=(1+\sin x)/\cos x. Squaring gives (1+sin⁡x)2/cos⁡2x(1+\sin x)^2/\cos^2x. Since cos⁡2x=(1−sin⁡x)(1+sin⁡x)\cos^2x=(1-\sin x)(1+\sin x), cancellation gives (1+sin⁡x)/(1−sin⁡x)(1+\sin x)/(1-\sin x) wherever the original expressions are defined. If sec⁡x+tan⁡x=3\sec x+\tan x=\sqrt3, the identity gives (1+sin⁡x)/(1−sin⁡x)=3(1+\sin x)/(1-\sin x)=3, so sin⁡x=1/2\sin x=1/2. The candidates in the interval are x=π/6x=\pi/6 and x=5π/6x=5\pi/6. Substitution into the unsquared equation gives 3\sqrt3 at π/6\pi/6 but −3-\sqrt3 at 5π/65\pi/6, so only x=π/6x=\pi/6 is valid.

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 5.8. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 5 Trigonometry

Want help turning this into marks?

Bring 5.8 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call