9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, 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 Solving Equations: 9. (a) Prove that sin 2x

9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan x sin x - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Question 2

9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan... show full transcript

Worked Solution & Example Answer:9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan x sin x - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Step 1

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

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Answer

To prove the identity, we start with the left side:

Use the double angle identity for sin: $sin 2x = 2 sin x cos x$ Therefore, $2 sin x cos x - tan x$ .
Substitute tan x with (\frac{sin x}{cos x}): $2 sin x cos x - \frac{sin x}{cos x}$ .
To combine the terms, find a common denominator (cos x): $\frac{2 sin x cos^2 x - sin x}{cos x}$ .
Factor out sin x: $sin x \left(2 cos^2 x - 1\right)$ .
Use the identity (cos 2x = 2 cos^2 x - 1): Hence, we can rewrite it as: $sin x \cdot cos 2x$ .
So, we have: $sin 2x - tan x = tan x cos 2x$ .

This proves the equation.

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:

We rewrite the left side: $sin 2x - tan x = 3 tan x sin x$ .
Substitute tan x using (\frac{sin x}{cos x}): $sin 2x - \frac{sin x}{cos x} = 3 \cdot \frac{sin x}{cos x} \cdot sin x$ .
Rearranging gives: $sin 2x = 4 \frac{sin^2 x}{cos x}$ .
Using the identity for sin 2x again: $2 sin x cos x = 4 \frac{sin^2 x}{cos x}$ .
Cross-multiplying leads to: $2 sin x cos^2 x = 4 sin^2 x$ .
Rearranging yields: $2 cos^2 x = 4 sin x$ .
Then divide by 2: $cos^2 x = 2 sin x$ .
Finally, express in terms of sin: $cos^2 x = 2(1 - cos^2 x)$ .
Rearranging gives: $3 cos^2 x + 2 = 0$ : This implies: (x = 16.3°, 163.7°, 180°).

Thus, the final solutions are: $x = 16.3°, 163.7°, 180°$ .

9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan x sin x - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

Worked Solution & Example Answer:9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan x sin x - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

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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