| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Prove using Pythagorean identity result |
| Difficulty | Moderate -0.3 Part (i) is a direct application of the Pythagorean identity sin²x + cos²x = 1, requiring simple algebraic expansion of (a+b)² + (a-b)² = 1 to get 2a² + 2b² = 1. Part (ii) uses tan x = sin x/cos x = 2 to create a linear equation in a and b. Both parts are straightforward with no novel insight required, making this slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a^2 + 2ab + b^2,\ a^2 - 2ab + b^2\) | B1 | Correct expansions |
| \(\sin^2 x + \cos^2 x = 1\) used \(\rightarrow (a+b)^2 + (a-b)^2 = 1\) | M1 | Appropriate use of \(\sin^2 x + \cos^2 x = 1\) with \((a+b)^2\) and \((a-b)^2\) |
| \(a^2 + b^2 = \frac{1}{2}\) | A1 | No evidence of \(\pm 2ab\), scores 2/3 |
| Alternative: \(2a = (s+c)\) & \(2b = (s-c)\) or \(a = \frac{1}{2}(s+c)\) & \(b = \frac{1}{2}(s-c)\) | B1 | |
| \(a^2+b^2 = \frac{1}{4}(s+c)^2 + \frac{1}{4}(s-c)^2 = \frac{1}{2}(s^2+c^2)\) | M1 | Appropriate use of \(\sin^2 x + \cos^2 x = 1\) |
| \(a^2 + b^2 = \frac{1}{2}\) | A1 | Method using only \((\sin x - b)^2\) and \((a - \cos x)^2\) scores 0/3. SC B1 for assuming \(\theta\) is acute giving \(a = \frac{1}{\sqrt{5}} + b\) or \(2\sqrt{5} - b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan x = \frac{\sin x}{\cos x} \rightarrow \frac{a+b}{a-b} = 2\) | M1 | Use of \(\tan x = \frac{\sin x}{\cos x}\) to form an equation in \(a\) and \(b\) only |
| \(a = 3b\) | A1 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a^2 + 2ab + b^2,\ a^2 - 2ab + b^2$ | B1 | Correct expansions |
| $\sin^2 x + \cos^2 x = 1$ used $\rightarrow (a+b)^2 + (a-b)^2 = 1$ | M1 | Appropriate use of $\sin^2 x + \cos^2 x = 1$ with $(a+b)^2$ and $(a-b)^2$ |
| $a^2 + b^2 = \frac{1}{2}$ | A1 | No evidence of $\pm 2ab$, scores 2/3 |
| **Alternative:** $2a = (s+c)$ & $2b = (s-c)$ or $a = \frac{1}{2}(s+c)$ & $b = \frac{1}{2}(s-c)$ | B1 | |
| $a^2+b^2 = \frac{1}{4}(s+c)^2 + \frac{1}{4}(s-c)^2 = \frac{1}{2}(s^2+c^2)$ | M1 | Appropriate use of $\sin^2 x + \cos^2 x = 1$ |
| $a^2 + b^2 = \frac{1}{2}$ | A1 | Method using only $(\sin x - b)^2$ and $(a - \cos x)^2$ scores 0/3. SC B1 for assuming $\theta$ is acute giving $a = \frac{1}{\sqrt{5}} + b$ or $2\sqrt{5} - b$ |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan x = \frac{\sin x}{\cos x} \rightarrow \frac{a+b}{a-b} = 2$ | M1 | Use of $\tan x = \frac{\sin x}{\cos x}$ to form an equation in $a$ and $b$ only |
| $a = 3b$ | A1 | |4 Angle $x$ is such that $\sin x = a + b$ and $\cos x = a - b$, where $a$ and $b$ are constants.\\
(i) Show that $a ^ { 2 } + b ^ { 2 }$ has a constant value for all values of $x$.\\
(ii) In the case where $\tan x = 2$, express $a$ in terms of $b$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q4 [5]}}