Problem 40 - Precalculus Essentials

Question

Verify each identity. $$\cot ^{2} 2 x+\cos ^{2} 2 x+\sin ^{2} 2 x=\csc ^{2} 2 x$$

Step-by-Step Solution

Answer

The given trigonometric identity is correct.

Step 1: Convert cotangent and cosecant in terms of sine and cosine

Use the identities \( \cot 2x = \frac{1}{\tan 2x}\) and \(\csc 2x = \frac{1}{\sin 2x} \) to write the given equation in terms of sine and cosine. After doing so, the equation will look like: \( \left(\frac{\cos 2x}{\sin 2x}\right)^2 + \sin^2 2x + \cos^2 2x = \left(\frac{1}{\sin 2x}\right)^2 \)

Step 2: Simplify

This will produce \( \frac{\cos^2 2x}{\sin^2 2x}+ \sin^2 2x + \cos^2 2x = \frac{1}{\sin^2 2x} \)or \( \frac{\cos^2 2x+ \sin^2 2x \sin^2 2x+ \sin^2 2x \cos^2 2x}{\sin^2 2x} = \frac{1}{\sin^2 2x}\)

Step 3: Apply Pythagorean identities

Notice that the left part of the equation can be reduced to \( \frac{1+\sin^4 2x+ \cos^4 2x}{\sin^2 2x} \) based on the Pythagorean identity \( \sin^2 x + \cos^2 x =1 \). The right side of the equation is \( \frac{1}{\sin^2 2x}\). The reduced equation is:\( \frac{1+\sin^4 2x+ \cos^4 2x}{\sin^2 2x} = \frac{1}{\sin^2 2x}\)

Step 4: Simplify again and reach final equation

If we subtract \( \frac{1}{\sin^2 2x}\) from both sides of the equation, we get \( \frac{1+\sin^4 2x + \cos^4 2x -1}{\sin^2 2x} = 0 \)which simplifies to \( \frac{\sin^4 2x + \cos^4 2x}{\sin^2 2x} = 0 \)This equation is always true, therefore the original identity holds true.