Question

Prove that (cos(3A) - sin(3A))/(1-2sin(2A)) ≡ cos(A) + sin(A)

Topic : Trigonometry
# Proving Trigonometric Identities
# Double Angle Identities
# Addition Identities

Answer 1

Answer:

Refer below.

Step-by-step explanation:

To address the trigonometric identity, we will approach the problem by simplifying each side of the equation using trigonometric identities and properties. Our goal is to transform one side of the equation into the other, or to show that both sides are equivalent through a series of logical steps.

We want to prove:

$\dfrac{\cos(3A) - \sin(3A)}{1-2\sin(2A)} \equiv \cos(A) + \sin(A)$

I always recommend simplifying the more complicated side, in this case the left-hand side. We start by expanding cos(3A) and sin(3A) using the triple angle formulas:

$\boxed{ \begin{array}{ccc} \text{\underline{Triple Angle Formulas:}} \\\\ \quad \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \\\\ \quad \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \\\\ \quad \tan(3\theta) = \dfrac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \end{array}}$

$\Longrightarrow \dfrac{\cos(3A) - \sin(3A)}{1-2\sin(2A)} \\\\\\\\\Longrightarrow \dfrac{[4\cos^3(A) - 3\cos(A)] - [3\sin(A) - 4\sin^3(A)]}{1-2\sin(2A)}$

Now applying the double-angle identity to the denominator:

$\boxed{ \begin{array}{ccc} \text{\underline{Double Angle Formulas:}} \\\\ \quad \sin(2\theta) = 2\sin(\theta)\cos(\theta) \\\\ \quad \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \\ \phantom{\text{Cosine:}} \quad \cos(2\theta) = 2\cos^2(\theta) - 1 \\ \phantom{\text{Cosine:}} \quad \cos(2\theta) = 1 - 2\sin^2(\theta) \\\\ \quad \tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^2(\theta)} \end{array}}$

$\Longrightarrow \dfrac{[4\cos^3(A) - 3\cos(A)] - [3\sin(A) - 4\sin^3(A)]}{1-2[2\sin(A)\cos(A)]}$

Rearranging this expression, we have:

$\Longrightarrow \dfrac{4\cos^3(A) - 3\cos(A)- 3\sin(\theta) + 4\sin^3(\theta)}{1-4\sin(A)\cos(A)}$

$\Longrightarrow \dfrac{ 4\sin^3(A)+4\cos^3(A)- 3\sin(A) - 3\cos(A)}{1-4\sin(A)\cos(A)}$

$\Longrightarrow \dfrac{ 4(\sin^3(A)+\cos^3(A))-3 (\sin(A) + \cos(A))}{1-4\sin(A)\cos(A)}$

Notice we have a sum of cubes in the numerator:

$\boxed{ \begin{array}{ccc} \text{\underline{Sum of Cubes Formula:}} \\\\ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \end{array}}$

$\Longrightarrow \dfrac{ 4\Big[\left(\sin(A)+\cos(A)\right)\left(\sin^2(A)-\sin(A)\cos(A)+\cos^2(A)\right)\Big] - 3 \Big[\sin(A) + \cos(A)\Big]}{1-4\sin(A)\cos(A)}$

Applying the pythagorean identity:

$\boxed{ \begin{array}{ccc} \text{\underline{Pythagorean Identities:}} \\\\ \sin^2(\theta) + \cos^2(\theta) = 1 \\\\ 1 + \tan^2(\theta) = \sec^2(\theta) \\\\ 1 + \cot^2(\theta) = \csc^2(\theta) \end{array}}$

$\Longrightarrow \dfrac{ 4\Big[\left(\sin(A)+\cos(A)\right)\left(1-\sin(A)\cos(A)\right)\Big] - 3 \Big[\sin(A) + \cos(A)\Big]}{1-4\sin(A)\cos(A)}$

$\Longrightarrow \dfrac{ 4\left(\sin(A)+\cos(A)\right)\left(1-\sin(A)\cos(A)\right) - 3 (\sin(A) + \cos(A))}{1-4\sin(A)\cos(A)}$

Factoring out 'sin(A) + cos(A)' in the numerator:

$\Longrightarrow \dfrac{ \left(\sin(A)+\cos(A)\right)\Big[4\left(1-\sin(A)\cos(A)\right) - 3 \Big]}{1-4\sin(A)\cos(A)}$

Simplifying this expression:

$\Longrightarrow \dfrac{ \left(\sin(A)+\cos(A)\right)\Big[4-4\sin(A)\cos(A)- 3 \Big]}{1-4\sin(A)\cos(A)}$

$\Longrightarrow \dfrac{ \left(\sin(A)+\cos(A)\right)\Big[1-4\sin(A)\cos(A) \Big]}{1-4\sin(A)\cos(A)}$

Canceling the common factor:

$\Longrightarrow \boxed{\sin(A)+\cos(A)}; \ \therefore \dfrac{\cos(3A) - \sin(3A)}{1-2\sin(2A)} \equiv \cos(A) + \sin(A)$

Thus, we have proved the identity.