| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity with double/compound angles |
| Difficulty | Moderate -0.3 Part (a) is a direct application of the compound angle formula with A=B=x, yielding the standard result tan(2x)=2tan(x)/(1-tan²(x)). Part (b) requires substituting this result and algebraic manipulation, but follows a clear path once the substitution is made. This is slightly easier than average as it's a guided proof with the identity provided and the result given to prove. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan(x+x) = \dfrac{\tan x + \tan x}{1 - \tan x \tan x} \left(= \dfrac{2\tan x}{1 - \tan^2 x}\right)\) | M1, A1 (2 marks) | \(A = B = x\) used |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 - 2\tan x - \dfrac{2\tan x(1-\tan^2 x)}{2\tan x}\) | M1 | Substitute from (a) |
| \(2 - 2\tan x - (1 - \tan x)(1 + \tan x)\) | M1 | Simplification \(2 - 2\tan x - (1 - \tan^2 x)\) |
| \((1 - \tan x)(2-(1+\tan x))\) | M1 | \(2 - 2\tan x - 1 + \tan^2 x\) |
| \((1-\tan x)^2\) | A1 (4 marks) | AG (convincingly obtained). \(= (\tan x - 1)^2 = (1-\tan x)^2\). Any equivalent method. |
# Question 7(a):
$\tan(x+x) = \dfrac{\tan x + \tan x}{1 - \tan x \tan x} \left(= \dfrac{2\tan x}{1 - \tan^2 x}\right)$ | M1, A1 (2 marks) | $A = B = x$ used
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# Question 7(b):
$2 - 2\tan x - \dfrac{2\tan x(1-\tan^2 x)}{2\tan x}$ | M1 | Substitute from (a)
$2 - 2\tan x - (1 - \tan x)(1 + \tan x)$ | M1 | Simplification $2 - 2\tan x - (1 - \tan^2 x)$
$(1 - \tan x)(2-(1+\tan x))$ | M1 | $2 - 2\tan x - 1 + \tan^2 x$
$(1-\tan x)^2$ | A1 (4 marks) | AG (convincingly obtained). $= (\tan x - 1)^2 = (1-\tan x)^2$. Any equivalent method.
---7
\begin{enumerate}[label=(\alph*)]
\item Use the identity
$$\tan ( A + B ) = \frac { \tan A + \tan B } { 1 - \tan A \tan B }$$
to express $\tan 2 x$ in terms of $\tan x$.
\item Show that
$$2 - 2 \tan x - \frac { 2 \tan x } { \tan 2 x } = ( 1 - \tan x ) ^ { 2 }$$
for all values of $x , \tan 2 x \neq 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q7 [6]}}