Standard +0.3 Part (a) is a standard derivation using tan(A+B) with A=B=x, then applying the formula again - a common textbook exercise. Part (b) requires algebraic manipulation to factor and solve a cubic in tan(x), which is routine once the identity is established. The question tests standard technique application rather than novel problem-solving, making it slightly easier than average.
8. (a) Using the trigonometric identity for \(\tan ( A + B )\), prove that
$$\tan 3 x = \frac { 3 \tan x - \tan ^ { 3 } x } { 1 - 3 \tan ^ { 2 } x } , \quad x \neq ( 2 n + 1 ) 30 ^ { \circ } , \quad n \in \mathbb { Z }$$
(b) Hence solve, for \(- 30 ^ { \circ } < x < 30 ^ { \circ }\),
$$\tan 3 x = 11 \tan x$$
(Solutions based entirely on graphical or numerical methods are not acceptable.)
Uses correct double angle formula \(\frac{2\tan x}{1-\tan^2 x}\) both times within expression for \(\tan(2x+x)\)
\(= \frac{2\tan x + \tan x(1-\tan^2 x)}{1-\tan^2 x - 2\tan^2 x}\) OR \(= \frac{\frac{2\tan x + \tan x - \tan^3 x}{1-\tan^2 x}}{\frac{1-\tan^2 x - 2\tan^2 x}{1-\tan^2 x}}\)
A1
Multiplies numerator and denominator by \(1-\tan^2 x\) to obtain correct intermediate line
Correct printed answer with no errors in all lines (c.s.o.). Withhold for poor notation e.g. \(\tan \leftrightarrow \tan x\), \(\tan^2 x \leftrightarrow \tan x^2\), \(\tan 2x = \frac{2\tan\theta}{1-\tan^2\theta}\)
(4)
Part (b):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
\(\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = 11\tan x\) so \(3\tan x - \tan^3 x = 11\tan x(1-3\tan^2 x)\)
M1
Attempts to use given identity and multiplies by \(1 - 3\tan^2 x\). Condone slips
\(32\tan^3 x = 8\tan x\)
A1
Obtain \(32\tan^3 x = 8\tan x\) or equivalent. Accept \(32\tan^2 x = 8\) for this mark
\(\tan x = \pm\frac{1}{2}\) or \(0 \Rightarrow x = \ldots\)
dM1
Obtains one value of \(x\) from \(\tan x = \ldots\) using correct method. Order of operations to find \(x\) must be correct but can be scored from \(\tan x = 0 \Rightarrow x = 0\)
\(x =\) awrt \(26.6°\), \(-26.6°\), \(0\)
A1 A1
A1: Either \(x = 26.6°\) or \(-26.6°\). A1: CAO \(x =\) awrt \(26.6°\), \(-26.6°\), \(0\) with no extras in range. Answers in radians \(\pm 0.46\)
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan(2x+x) = \frac{\tan 2x + \tan x}{1 - \tan 2x\tan x}$ | M1 | Expands $\tan(2x+x) = \frac{\tan 2x + \tan x}{1 - \tan 2x\tan x}$, condoning sign errors |
| $= \frac{\frac{2\tan x}{1-\tan^2 x} + \tan x}{1 - \frac{2\tan x}{1-\tan^2 x}\tan x}$ | dM1 | Uses correct double angle formula $\frac{2\tan x}{1-\tan^2 x}$ both times within expression for $\tan(2x+x)$ |
| $= \frac{2\tan x + \tan x(1-\tan^2 x)}{1-\tan^2 x - 2\tan^2 x}$ OR $= \frac{\frac{2\tan x + \tan x - \tan^3 x}{1-\tan^2 x}}{\frac{1-\tan^2 x - 2\tan^2 x}{1-\tan^2 x}}$ | A1 | Multiplies numerator and denominator by $1-\tan^2 x$ to obtain correct intermediate line |
| $\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$ * | A1* cso | Correct printed answer with no errors in all lines (c.s.o.). Withhold for poor notation e.g. $\tan \leftrightarrow \tan x$, $\tan^2 x \leftrightarrow \tan x^2$, $\tan 2x = \frac{2\tan\theta}{1-\tan^2\theta}$ |
| | **(4)** | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = 11\tan x$ so $3\tan x - \tan^3 x = 11\tan x(1-3\tan^2 x)$ | M1 | Attempts to use given identity and multiplies by $1 - 3\tan^2 x$. Condone slips |
| $32\tan^3 x = 8\tan x$ | A1 | Obtain $32\tan^3 x = 8\tan x$ or equivalent. Accept $32\tan^2 x = 8$ for this mark |
| $\tan x = \pm\frac{1}{2}$ or $0 \Rightarrow x = \ldots$ | dM1 | Obtains one value of $x$ from $\tan x = \ldots$ using correct method. Order of operations to find $x$ must be correct but can be scored from $\tan x = 0 \Rightarrow x = 0$ |
| $x =$ awrt $26.6°$, $-26.6°$, $0$ | A1 A1 | A1: Either $x = 26.6°$ or $-26.6°$. A1: CAO $x =$ awrt $26.6°$, $-26.6°$, $0$ with no extras in range. Answers in radians $\pm 0.46$ |
| | **(5)** | Answers only scores 0. Correct cubic/quadratic scores M1 A1 dM1 (implied) |
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8. (a) Using the trigonometric identity for $\tan ( A + B )$, prove that
$$\tan 3 x = \frac { 3 \tan x - \tan ^ { 3 } x } { 1 - 3 \tan ^ { 2 } x } , \quad x \neq ( 2 n + 1 ) 30 ^ { \circ } , \quad n \in \mathbb { Z }$$
(b) Hence solve, for $- 30 ^ { \circ } < x < 30 ^ { \circ }$,
$$\tan 3 x = 11 \tan x$$
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C34 2017 Q8 [9]}}