[ sin 2pi/9 + icos 2pi/9 + 1 / sin 2pi/9 - icos 2pi/9 + 1 ]^3 =

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – AP-EAMCET 2026

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Home Maths Complex & QuadraticQuestion #59

{[\frac{\sin \frac{2 π}{9} + i \cos \frac{2 π}{9} + 1}{\sin \frac{2 π}{9} - i \cos \frac{2 π}{9} + 1}]}^{3}

Clue 1 of 3: Complex Number Simplification

Recognize the form of the expression involving trigonometric functions and consider converting them to exponential form to simplify the structure.

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Step 1: Rewrite Trigonometric Terms in Euler's Form✦ Active

Let $θ = \frac{2 π}{9}$ . The given expression is ${[\frac{1 + \sin θ + i \cos θ}{1 + \sin θ - i \cos θ}]}^{3}$ . We can rewrite the terms $\sin θ + i \cos θ$ and $\sin θ - i \cos θ$ using Euler's formula. First, express $\sin θ$ as $\cos (\frac{π}{2} - θ)$ and $\cos θ$ as $\sin (\frac{π}{2} - θ)$ . Let $α = \frac{π}{2} - θ$ .

\sin θ + i \cos θ = \cos (\frac{π}{2} - θ) + i \sin (\frac{π}{2} - θ) = e^{i (\frac{π}{2} - θ)} = e^{i α}

Similarly, the conjugate term is:

\sin θ - i \cos θ = \cos (\frac{π}{2} - θ) - i \sin (\frac{π}{2} - θ) = e^{- i (\frac{π}{2} - θ)} = e^{- i α}

Substitute $θ = \frac{2 π}{9}$ to find $α$ :

α = \frac{π}{2} - \frac{2 π}{9} = \frac{9 π - 4 π}{18} = \frac{5 π}{18}

💡 Teacher's Secret Hint

Remember that $\sin x + i \cos x$ is not directly $e^{i x}$ , but it can be transformed by swapping sine and cosine using complementary angles, i.e., $\cos (\frac{π}{2} - x) + i \sin (\frac{π}{2} - x)$ .

Step 2: Simplify the Fraction Inside the Bracket○ Expand

The expression inside the bracket becomes $\frac{1 + e^{i α}}{1 + e^{- i α}}$ . We use the identity $1 + e^{i ϕ} = e^{i ϕ / 2} (e^{- i ϕ / 2} + e^{i ϕ / 2}) = 2 \cos (ϕ / 2) e^{i ϕ / 2}$ .

\frac{1 + e^{i α}}{1 + e^{- i α}} = \frac{2 \cos (α / 2) e^{i α / 2}}{2 \cos (α / 2) e^{- i α / 2}}

Since $α = \frac{5 π}{18}$ , $α / 2 = \frac{5 π}{36}$ . $\cos (\frac{5 π}{36}) \neq 0$ , so we can cancel the $\cos (α / 2)$ terms.

\frac{e^{i α / 2}}{e^{- i α / 2}} = e^{i α / 2} \cdot e^{i α / 2} = e^{i α}

💡 Teacher's Secret Hint

This identity is very useful for simplifying expressions of the form $1 + e^{i ϕ}$ or $1 - e^{i ϕ}$ . Factoring out $e^{i ϕ / 2}$ often simplifies the expression significantly.

Step 3: Apply De Moivre's Theorem○ Expand

Now, substitute $e^{i α}$ back into the original expression and apply the power of 3.

{[e^{i α}]}^{3} = e^{i 3 α}

Substitute the value of $α = \frac{5 π}{18}$ .

e^{i 3 (\frac{5 π}{18})} = e^{i \frac{5 π}{6}}

Step 4: Convert to Rectangular Form○ Expand

Convert the final exponential form back to rectangular form using Euler's formula $e^{i ϕ} = \cos ϕ + i \sin ϕ$ .

e^{i \frac{5 π}{6}} = \cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6})

Evaluate the trigonometric values:

\cos (\frac{5 π}{6}) = \cos (π - \frac{π}{6}) = - \cos (\frac{π}{6}) = - \frac{\sqrt{3}}{2}

\sin (\frac{5 π}{6}) = \sin (π - \frac{π}{6}) = \sin (\frac{π}{6}) = \frac{1}{2}

Combine these values:

e^{i \frac{5 π}{6}} = - \frac{\sqrt{3}}{2} + i \frac{1}{2} = - \frac{1}{2} (\sqrt{3} - i)

💡 Teacher's Secret Hint

Always check the quadrant of the angle when evaluating sine and cosine to ensure the correct sign. $\frac{5 π}{6}$ is in the second quadrant where cosine is negative and sine is positive.

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Maths Question 9 – AP-EAMCET 2026

Clue 1 of 3: Complex Number Simplification

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