| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Standard +0.3 This is a structured three-part question testing standard trigonometric identities. Part (i) requires proving a given identity using basic definitions (tan = sin/cos, sec = 1/cos), part (ii) is a straightforward substitution (β = α), and part (iii) involves solving a quadratic in tan θ. All techniques are routine C4 content with clear signposting ('show that', 'hence'), making it slightly easier than average but still requiring careful algebraic manipulation. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | EITHER Use of cos = 1/sec (or sin= 1/cosec) |
| Answer | Marks |
|---|---|
| sec sec | B1 |
| Answer | Marks |
|---|---|
| [3] | Must be used |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | 1 tan2 |
| Answer | Marks |
|---|---|
| . | M1 |
| A1 | β = α used , Need to see sec²α |
| Answer | Marks |
|---|---|
| 1 tan | M1 |
| [2] | Use of cos2α = cos²α – sin²α soi |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (iii) | cos 2 = ½ |
| Answer | Marks |
|---|---|
| = 30°, 150° | M1 |
| Answer | Marks |
|---|---|
| [3] | Soi or from tan²θ = 1/3 oe from sin²θ or cos²θ |
Question 7:
7 | (i) | EITHER Use of cos = 1/sec (or sin= 1/cosec)
From RHS
1 tan tan
sec sec
1 sin /cos .sin /cos
1/cos .1/cos
sin sin
cos cos (1 )
cos cos
cos cos sin sin
cos( )
OR From LHS, cos = 1/sec or sin = 1/cosec used
cos( )
cos cos sin sin
1
sin sin
sec sec
1 sec sin sec sin
sec sec
1 tan tan
sec sec | B1
M1
A1
B1
M1
A1
[3] | Must be used
Substituting and simplifying as far as having no fractions within a
fraction
1 tt
[need more than cc ss ie an intermediate step that can lead
secsec
to cc–ss]
Convincing simplification and correct use of cos(α + β)
Answer given
Correct angle formula and substitution and simplification to one term
OR eg cosαcosβ – sinαsinβ
cos cos (1 tan tan )
Simplifying to final answer www
Answer given
Or any equivalent work but must have more than cc–ss = answer.
7 | (ii) | 1 tan2
cos2
sec2
1 tan2
1 tan2
. | M1
A1 | β = α used , Need to see sec²α
Use of sec²α = 1 + tan²α to give required result
Answer Given
OR, without Hence,
sin2
cos2 cos2 (1 )
cos2
1
(1 tan2 )
sec2
1 tan2
2
1 tan | M1
[2] | Use of cos2α = cos²α – sin²α soi
Simplifying and using sec²α = 1+ tan²α to final answer
Answer Given
Accept working in reverse to show RHS=LHS, or showing equivalent
7 | (iii) | cos 2 = ½
2 = 60°, 300°
= 30°, 150° | M1
A1
A1
[3] | Soi or from tan²θ = 1/3 oe from sin²θ or cos²θ
First correct solution
Second correct solution and no others in the range
SC B1 for π/6and 5π/6 and no others in the range\begin{enumerate}[label=(\roman*)]
\item Show that $\cos(\alpha + \beta) = \frac{1 - \tan\alpha\tan\beta}{\sec\alpha\sec\beta}$. [3]
\item Hence show that $\cos 2\alpha = \frac{1 - \tan^2\alpha}{1 + \tan^2\alpha}$. [2]
\item Hence or otherwise solve the equation $\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1}{2}$ for $0° \leqslant \theta \leqslant 180°$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q7 [8]}}