Question

Solve the following trigonometric expression
$\sin \theta + \sin 2\theta + \sin 3\theta + \sin 4\theta = 0$.

Answer 1

Hint: In the questions where angles are given in terms of $\theta$, we need to find out the formulas that can be used to solve these. After simplifying the giving equation use the properties of sinθ and cosθ and then proceed further.

Complete step-by-step answer:

$(\sin 4x + \sin x) + (\sin 3x + \sin 2x) = 0$

Using the formula
$\sin \alpha + \sin \beta = 2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$

So,$2\sin \left( {\dfrac{{4x + x}}{2}} \right)\cos \left( {\dfrac{{4x - x}}{2}} \right) + 2\sin \left( {\dfrac{{3x + 2x}}{2}} \right)\cos \left( {\dfrac{{3x - 2x}}{2}} \right) = 0$
$\Rightarrow 2\sin \left( {\dfrac{5}{2}x} \right)\cos \left( {\dfrac{3}{2}x} \right) + 2\sin \left( {\dfrac{5}{2}x} \right)\cos \left( {\dfrac{1}{2}x} \right) = 0$
$\Rightarrow 2\sin \left( {\dfrac{5}{2}x} \right)\left[ {\cos \left( {\dfrac{3}{2}x} \right) + \cos \left( {\dfrac{1}{2}x} \right)} \right] = 0$

Now using
$\cos \alpha + \cos \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right)$ we get
$\Rightarrow 2\sin \left( {\dfrac{5}{2}x} \right)2\cos \left( {\dfrac{{\dfrac{3}{2}x + \dfrac{1}{2}x}}{2}} \right)\cos \left( {\dfrac{{\dfrac{3}{2}x - \dfrac{1}{2}x}}{2}} \right) = 0$
$\Rightarrow 4\sin \left( {\dfrac{5}{2}x} \right)\cos x\cos \left( {\dfrac{x}{2}} \right) = 0$

Then:
$\Rightarrow \sin \left( {\dfrac{5}{2}x} \right) = 0 \Rightarrow \dfrac{5}{2}x = K\pi$
$\Rightarrow x = \dfrac{2}{5}k\pi$

Similarly,
$\Rightarrow \cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi$
∴ $\cos \left( {\dfrac{x}{2}} \right) = 0$
$\Rightarrow \dfrac{x}{2} = \dfrac{\pi }{2} + k\pi$
$\Rightarrow x = \pi + 2k\pi$

Note: Following are the properties of sinθ and cosθ. While solving using these formulas don’t be confused with - and + signs (Major mistakes are made by this confusion). A good command over trigonometric identities and formulas will be an added advantage.

$$\sin \alpha + \sin \beta = 2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) \\\ \cos \alpha + \cos \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) \\\ \sin \alpha - \sin \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right) \\\ \cos \alpha - \cos \beta = - 2\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right) \\\$$