| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Prove Pythagorean identity from triangle |
| Difficulty | Easy -1.2 This is a foundational proof requiring only basic Pythagoras theorem application and simple algebraic manipulation of the Pythagorean identity. Part (i) is direct recall from a right-angled triangle, and part (ii) follows immediately by dividing through by cos²x and using tan x = sin x/cos x. This is easier than average as it's a standard textbook derivation with no problem-solving required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| In rt-angled triangle, \(a^2+b^2=c^2 \Rightarrow \left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1\) | M1 | |
| \(\sin x = \frac{a}{c},\ \cos x = \frac{b}{c} \Rightarrow \sin^2 x + \cos^2 x = 1\) | M1 A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Divide by \(\cos^2 x\): \(\frac{\sin^2 x}{\cos^2 x}+1 = \frac{1}{\cos^2 x} \Rightarrow \tan^2 x + 1 = \frac{1}{\cos^2 x}\) | M1 A1 | Total: 2 |
## Question 7(i):
In rt-angled triangle, $a^2+b^2=c^2 \Rightarrow \left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$ | M1 |
$\sin x = \frac{a}{c},\ \cos x = \frac{b}{c} \Rightarrow \sin^2 x + \cos^2 x = 1$ | M1 A1 | **Total: 3**
## Question 7(ii):
Divide by $\cos^2 x$: $\frac{\sin^2 x}{\cos^2 x}+1 = \frac{1}{\cos^2 x} \Rightarrow \tan^2 x + 1 = \frac{1}{\cos^2 x}$ | M1 A1 | **Total: 2**
---7 (i) Using the triangle, show that $\sin ^ { 2 } x + \cos ^ { 2 } x = 1$.\\
(ii) Hence prove that\\
\includegraphics[max width=\textwidth, alt={}]{73d1c02b-1b7b-426d-a171-c762597cfed4-2_255_501_1779_1022} $1 + \tan ^ { 2 } x = \frac { 1 } { \cos ^ { 2 } x }$.
\hfill \mbox{\textit{OCR MEI C2 Q7 [5]}}