f(x) = 7 cos 2x - 24 sin 2x Given that f(x) = R cos(2x + α), where R > 0 and 0 < α < 90°; (a) find the value of R and the value of α - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 5

Substituting the value of R and α from part (a), we rewrite the equation as:

$25 cos(2x + 73.7°) = 12.5$

Dividing both sides by 25:

$cos(2x + 73.7°) = \frac{12.5}{25} = 0.5$

Thus:

$2x + 73.7° = 60°\ ext{ or }\ 2x + 73.7° = 300°$

Solving these gives:

1. For ( 2x + 73.7° = 60°:
   2x = 60° - 73.7°
   2x = -13.7°
   x = -6.85°\ (not in range)\

2. For ( 2x + 73.7° = 300° :
   2x = 300° - 73.7° \
   2x = 226.3°\
   x = 113.15°\

3. Since the cosine function has a periodicity of 360°, we find:

We start by using the double angle formulas:

1. Rewrite $cos² x$: $cos² x = \frac{1 + cos(2x)}{2}$

2. Rewrite $sin x cos x$: $sin x cos x = \frac{1}{2} sin(2x)$

Plugging these into the expression gives:

$14 cos² x - 48 sin x cos x = 14 \left( \frac{1 + cos(2x)}{2} \right) - 48 \left( \frac{1}{2} sin(2x) \right)$

This simplifies to:

$= 7 + 7 cos(2x) - 24 sin(2x)$

Comparing this with the form $a cos(2x) + b sin(2x) + c$, we find:

• a = 7
• b = -24
• c = 7.

The maximum value of an expression of the form $R cos(θ) + b sin(θ) + c$ can be determined as follows:

From part (c), we rewrite the expression as:

a cos(2x) + b sin(2x) + c, where

• a = 7
• b = -24
• c = 7

The maximum value of the term $a cos(θ) + b sin(θ)$ is given by:

$\sqrt{a^2 + b^2} = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$

Therefore, the maximum value of ( 14 cos^2 x - 48 sin x cos x ) is:

$max = 25 + c = 25 + 7 = 32$