9. (a) Prove that $$ \sin 2x - \tan x = \tan x \cos 2x, \quad x \neq (2n + 1)90^{\circ}, \quad n \in \mathbb{Z} $$ (b) Given that $x \neq 90^{\circ}$ and $x \neq 270^{\circ}$, solve, for $0 \leq x < 360^{\circ}$, $$ \sin 2x - \tan x = 3\tan x \sin x. $$ Give your answers in degrees to one decimal place where appropriate. (Solutions based entirely on graphical or numerical methods are not acceptable.)

A-Level Edexcel Maths Pure Differentiation: 9. (a) Prove that $$ \sin 2x -

9. (a) Prove that $$ \sin 2x - \tan x = \tan x \cos 2x, \quad x \neq (2n + 1)90^{\circ}, \quad n \in \mathbb{Z} $$ (b) Given that $x \neq 90^{\circ}$ and $x \neq 270^{\circ}$, solve, for $0 \leq x < 360^{\circ}$, $$ \sin 2x - \tan x = 3\tan x \sin x - Edexcel - A-Level Maths Pure - Question 2 - 2016 - Paper 3

Question 2

$9.-(a)-Prove-that--$$-\sin-2x---\tan-x-=-\tan-x-\cos-2x,-\quad-x-\neq-(2n-+-1)90^{\circ},-\quad-n-\in-\mathbb{Z}-$$--(b)-Given-that-$x-\neq-90^{\circ}$-and-$x-\neq-270^{\circ}$,-solve,-for-$0-\leq-x-<-360^{\circ}$,--$$-\sin-2x---\tan-x-=-3\tan-x-\sin-x-Edexcel-A-Level Maths Pure-Question 2-2016-Paper 3.png$

9. (a) Prove that $$ \sin 2x - \tan x = \tan x \cos 2x, \quad x \neq (2n + 1)90^{\circ}, \quad n \in \mathbb{Z} $$ (b) Given that $x \neq 90^{\circ}$ and \(x \ne... show full transcript

Worked Solution & Example Answer:9. (a) Prove that $$ \sin 2x - \tan x = \tan x \cos 2x, \quad x \neq (2n + 1)90^{\circ}, \quad n \in \mathbb{Z} $$ (b) Given that $x \neq 90^{\circ}$ and $x \neq 270^{\circ}$, solve, for $0 \leq x < 360^{\circ}$, $$ \sin 2x - \tan x = 3\tan x \sin x - Edexcel - A-Level Maths Pure - Question 2 - 2016 - Paper 3

Step 1

Prove that sin 2x - tan x = tan x cos 2x

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Answer

To prove the equation, we start from the left-hand side:

\sin 2x - \tan x = \sin 2x - \frac{\sin x}{\cos x}.

Using the double angle identity for sine, we know:

\sin 2x = 2\sin x \cos x.

Thus, substituting this into our equation gives:

2\sin x \cos x - \frac{\sin x}{\cos x}.

Finding a common denominator:

= \frac{2\sin x \cos^2 x - \sin x}{\cos x}.

Factoring out (\sin x):

= \frac{\sin x (2\cos^2 x - 1)}{\cos x}.

Applying the identity (2\cos^2 x - 1 = \cos 2x):

= \frac{\sin x \cos 2x}{\cos x} = \tan x \cos 2x.

Thus, we have proved that:

\sin 2x - \tan x = \tan x \cos 2x.

This completes the proof.

Step 2

Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°: sin 2x - tan x = 3 tan x sin x

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Answer

Starting with the equation:

\sin 2x - \tan x = 3\tan x \sin x.

Substituting the double angle identity for sine:

2\sin x \cos x - \tan x = 3\tan x \sin x.

This means:

2\sin x \cos x - \frac{\sin x}{\cos x} = 3\cdot\frac{\sin x}{\cos x}\cdot \sin x.

Multiplying through by (\cos^2 x) (valid since (x \neq 90^{\circ}) and (x \neq 270^{\circ})) gives:

2\sin x \cos^3 x - \sin x = 3\sin^2 x \cos x.

Rearranging this leads to:

2\sin x \cos^3 x - 3\sin^2 x \cos x - \sin x = 0.

Factoring out (\sin x":

\sin x (2\cos^3 x - 3\sin x \cos x - 1) = 0.

This results in:

(\sin x = 0) gives (x = 0°, 180°);
Solving (2\cos^3 x - 3\sin x \cos x - 1 = 0) leads to:

Using a numerical or analytical approach, it can be shown:

One solution is approximately (16.3°);
Another is approximately (163.7°); Thus the solutions are:

$x = 0°, 16.3°, 163.7°, 180°.\$

These values are within the required range.

Prove that sin 2x - tan x = tan x cos 2x

Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°: sin 2x - tan x = 3 tan x sin x

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