How to write 2sinθ - 4cosθ in the form rsin(θ - α)?

Equating parts,

To write (2\sin\theta - 4\cos\theta) in the form (r\sin(\theta - \alpha)), you can use the following steps:

1. Use the trigonometric identity: (r\sin(\theta - \alpha) = r(\sin\theta\cos\alpha - \cos\theta\sin\alpha)).
2. Compare the expression (2\sin\theta - 4\cos\theta) with (r(\sin\theta\cos\alpha - \cos\theta\sin\alpha)).
3. Equate the coefficients of (\sin\theta) and (\cos\theta) in both expressions to find (r) and (\alpha).

Let's proceed with the calculations:

Given expression: (2\sin\theta - 4\cos\theta)

Comparing with (r(\sin\theta\cos\alpha - \cos\theta\sin\alpha)):

Coefficient of (\sin\theta): (2 = r\cos\alpha)

Coefficient of (\cos\theta): (-4 = -r\sin\alpha)

From the first equation, solve for (r): (r = \frac{2}{\cos\alpha})

From the second equation, solve for (\alpha): (\sin\alpha = \frac{4}{r} = \frac{4}{\frac{2}{\cos\alpha}} = \frac{4\cos\alpha}{2} = 2\cos\alpha)

Now, square both sides: (\sin^2\alpha = 4\cos^2\alpha)

Use the identity (\sin^2\alpha + \cos^2\alpha = 1): (1 - \cos^2\alpha = 4\cos^2\alpha)

Rearrange and solve for (\cos^2\alpha): (5\cos^2\alpha = 1) (\cos^2\alpha = \frac{1}{5})

Take the square root: (\cos\alpha = \pm \frac{1}{\sqrt{5}})

Since (r = \frac{2}{\cos\alpha}), and (\cos\alpha) cannot be negative for (0 < \theta < 360^\circ), we take the positive value for (\cos\alpha).

(\cos\alpha = \frac{1}{\sqrt{5}})

Now, substitute the value of (r) into the equation: (r = \frac{2}{\frac{1}{\sqrt{5}}} = 2\sqrt{5})

So, (r = 2\sqrt{5}) and (\alpha = \arccos\left(\frac{1}{\sqrt{5}}\right)).

Therefore, (2\sin\theta - 4\cos\theta = 2\sqrt{5}\sin\left(\theta - \arccos\left(\frac{1}{\sqrt{5}}\right)\right)).