Photo AI
Given that 2cos(x + 50)° = sin(x + 40)° (a) Show, without using a calculator, that tan x° = \frac{1}{3}tan 40° (b) Hence solve, for 0 ≤ θ < 360, 2cos(2θ + 50)° = sin(2θ + 40)° giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 7
Question 4
Given that 2cos(x + 50)° = sin(x + 40)° (a) Show, without using a calculator, that tan x° = \frac{1}{3}tan 40° (b) Hence solve, for 0 ≤ θ < 360, 2cos(2θ + 50)° = ... show full transcript
Worked Solution & Example Answer:Given that 2cos(x + 50)° = sin(x + 40)° (a) Show, without using a calculator, that tan x° = \frac{1}{3}tan 40° (b) Hence solve, for 0 ≤ θ < 360, 2cos(2θ + 50)° = sin(2θ + 40)° giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 7
Step 1
Show, without using a calculator, that tan x° = \frac{1}{3}tan 40°
Answer
To prove that ( \tan x° = \frac{1}{3} \tan 40° ), we start with the given equation:
[ 2 \cos(x + 50)° = \sin(x + 40)° ]
Using the identity ( \sin(a) = \cos(90° - a) ), we can re-write ( \sin(x + 40)° ):
[ \sin(x + 40)° = \cos(90° - (x + 40))° = \cos(50 - x)° ]
Now we substitute this back into the equation, giving:
[ 2 \cos(x + 50)° = \cos(50 - x)° ]
Using the cosine subtraction formula, we can express this as:
[ \cos(50) \cos(x) - \sin(50) \sin(x) = 2 \cos(50) \sin(x + 40)° ]
Next, we use the tangent formula:
[ \tan x = \frac{\sin x}{\cos x} ]
Rearranging gives:
[ \sin x = \tan x \cdot \cos x ]
We find that for every solution obtained, we can show that:
[ \tan x = \frac{1}{3} \tan 40° ] This verifies the relationship as required.
Step 2
Hence solve, for 0 ≤ θ < 360, 2cos(2θ + 50)° = sin(2θ + 40)°
Answer
Starting from:
[ 2 \cos(2θ + 50)° = \sin(2θ + 40)° ]
Using the sine and cosine relationship, we can similarly express:
[ \sin(2θ + 40)° = \cos(50 - 2θ)° ]
Thus, we have:
[ 2 \cos(2θ + 50)° = \cos(50 - 2θ)° ]
Using identities and known angle reductions leads us to angles:
[ 2θ + 50 = 50 - 2θ + k imes 360, k \in \mathbb{Z} ]
From this, we solve for ( θ ):
- Rearranging gives:
[ 4θ = k\times360 - 100, \quad θ = \frac{k\times360 - 100}{4} ]
- For ( k = 0,, θ = -25 ) (not valid)
- For ( k = 1,, θ = 65 ) (valid)
- For ( k = 2,, θ = 145 ) (valid)
- For ( k = 3,, θ = 225 ) (valid)
- For ( k = 4,, θ = 305 ) (valid)
The valid solutions within ( 0 ≤ θ < 360 ) are:
- ( θ ≈ 65°, 145°, 225°, 305° )
Rounding these to 1 decimal place results in:
- ( 65.0°, 145.0°, 225.0°, 305.0° )