Solve the trigonometric problem: If cos X + cos² X = 1, find sin¹² X + 3 sin¹⁰ X + 3 sin⁸ X + sin⁶ X - 1. Detailed solution

MHT-CET Trigonometric Problem Solution

Question: If \( \cos X + \cos^2 X = 1 \), then find the value of \( \sin^{12} X + 3 \sin^{10} X + 3 \sin^{8} X + \sin^{6} X - 1 \).

Options:

a) 2

b) 1

c) -1

d) 0

Detailed Solution

Let’s solve this step-by-step in a way that’s easy to follow. We’re given the equation \( \cos X + \cos^2 X = 1 \), and we need to find the value of the expression \( \sin^{12} X + 3 \sin^{10} X + 3 \sin^{8} X + \sin^{6} X - 1 \). The options are 2, 1, -1, and 0.

Step 1: Solve for \( \cos X \)

Start with the given equation:

\[ \cos X + \cos^2 X = 1 \]

Rearrange it to form a quadratic equation in terms of \( \cos X \). Let’s define \( y = \cos X \):

\[ y^2 + y - 1 = 0 \]

Use the quadratic formula to solve for \( y \), where the formula is:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \):

\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

So, the solutions are:

\[ y = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad y = \frac{-1 - \sqrt{5}}{2} \]

Since \( \cos X \) must be between -1 and 1, check each value:

• \( \frac{-1 + \sqrt{5}}{2} \): Knowing \( \sqrt{5} \approx 2.236 \), compute \( -1 + 2.236 = 1.236 \), then \( \frac{1.236}{2} \approx 0.618 \). This is between -1 and 1, so it’s valid.
• \( \frac{-1 - \sqrt{5}}{2} \): Compute \( -1 - 2.236 = -3.236 \), then \( \frac{-3.236}{2} \approx -1.618 \). This is less than -1, so it’s not possible for \( \cos X \).

Thus, \( \cos X = \frac{-1 + \sqrt{5}}{2} \).

Step 2: Find \( \sin^2 X \)

Use the Pythagorean identity:

\[ \sin^2 X + \cos^2 X = 1 \] \[ \sin^2 X = 1 - \cos^2 X \]

First, calculate \( \cos^2 X \):

\[ \cos X = \frac{-1 + \sqrt{5}}{2} \] \[ \cos^2 X = \left( \frac{-1 + \sqrt{5}}{2} \right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]

Now, compute \( \sin^2 X \):

\[ \sin^2 X = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2}{2} - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{2 - 3 + \sqrt{5}}{2} = \frac{-1 + \sqrt{5}}{2} \]

Notice that \( \sin^2 X = \frac{-1 + \sqrt{5}}{2} \), which equals \( \cos X \). So:

\[ \sin^2 X = \cos X \]

Step 3: Simplify the Expression

We need to evaluate:

\[ \sin^{12} X + 3 \sin^{10} X + 3 \sin^{8} X + \sin^{6} X - 1 \]

The coefficients 1, 3, 3, 1 and the powers 12, 10, 8, 6 suggest a binomial pattern. Rewrite the terms using \( \sin^2 X \):

\[ \sin^{12} X = (\sin^2 X)^6, \quad \sin^{10} X = (\sin^2 X)^5, \quad \sin^{8} X = (\sin^2 X)^4, \quad \sin^{6} X = (\sin^2 X)^3 \]

The expression resembles \( (a + b)^3 \). Test if it fits:

\[ (a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 \]

Let \( a = \sin^4 X \) and \( b = \sin^2 X \):

\[ ( \sin^4 X + \sin^2 X )^3 = (\sin^4 X)^3 + 3 (\sin^4 X)^2 (\sin^2 X) + 3 (\sin^4 X) (\sin^2 X)^2 + (\sin^2 X)^3 \] \[ = \sin^{12} X + 3 \sin^8 X \sin^2 X + 3 \sin^4 X \sin^4 X + \sin^6 X \] \[ = \sin^{12} X + 3 \sin^{10} X + 3 \sin^8 X + \sin^6 X \]

This matches our expression without the -1. So:

\[ \sin^{12} X + 3 \sin^{10} X + 3 \sin^8 X + \sin^6 X = ( \sin^4 X + \sin^2 X )^3 \] \[ \sin^{12} X + 3 \sin^{10} X + 3 \sin^8 X + \sin^6 X - 1 = ( \sin^4 X + \sin^2 X )^3 - 1 \]

Step 4: Substitute Using \( \sin^2 X = \cos X \)

Since \( \sin^2 X = \cos X \):

\[ \sin^4 X = (\sin^2 X)^2 = (\cos X)^2 = \cos^2 X \] \[ \sin^4 X + \sin^2 X = \cos^2 X + \cos X \]

From the original equation:

\[ \cos X + \cos^2 X = 1 \] \[ \cos^2 X + \cos X = 1 \]

So:

\[ \sin^4 X + \sin^2 X = \cos^2 X + \cos X = 1 \] \[ ( \sin^4 X + \sin^2 X )^3 = 1^3 = 1 \] \[ ( \sin^4 X + \sin^2 X )^3 - 1 = 1 - 1 = 0 \]

Step 5: Verify

The expression simplifies to 0, which is option d. This solution leverages the given equation directly, making it efficient and clear.

Final Value: 0