g(θ) = 4cos 2θ + 2sin 2θ Given that g(θ) = R cos(2θ - α), where R > 0 and 0 < α < 90°, (a) find the exact value of R and the value of α to 2 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 3

Using the previous result we substitute:

$R\cos(2θ - 26.57°) = 1$

This gives us:

$2\text{sqrt}(5)\cos(2θ - 26.57°) = 1$

Solving for cos, we have:

$\cos(2θ - 26.57°) = \frac{1}{2\text{sqrt}(5)}$

Now, we find the values of (2θ):

$\cos(2θ - 26.57°) = 0.4472$

Then,

$2θ - 26.57° = \pm\arccos(0.4472)$

Calculating the two cases will provide the general solutions. We can calculate that:

1. For the principal value: (2θ - 26.57° \approx 63.43° \Rightarrow θ \approx 45.71° )
2. For the secondary value: (2θ - 26.57° = -63.43° \Rightarrow θ \approx -45.71° )

Now we constrain these solutions to the range −90° < θ < 90° which gives us: Final Answers: (θ ext{ (to one decimal place)} \approx 45.7° ext{ and } -45.7°.

The equation g(θ) = k has no solutions when k is outside the range of values that g(θ) can achieve. We first find the maximum and minimum values of g(θ). Since g(θ) = R cos(2θ - α), the maximum value of g(θ) occurs when:\n1. (R = 2\text{sqrt}(5))\n2. (cos(2θ - α) = 1)\n Thus, maximum value of g(θ) = $2\text{sqrt}(5)$, and minimum occurs:\n1. (cos(2θ - α) = -1)\nThus, the minimum value = -$2\text{sqrt}(5)$. Hence, the range of k for no solutions is:\n Range: ( k < -2\text{sqrt}(5) \text{ or } k > 2\text{sqrt}(5) \approx \pm 4.472)