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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

Use De Moivre theorem to find cos 3θ in terms of cos θ.

Answer: 4cos³θ - 3cosθ.

  • A 3cosθ - 4cos³θ
  • B cos³θ - sin²θ
  • C 3cos²θ - 1
  • D 4cos³θ - 3cosθ

Correct answer: D. 4cos³θ - 3cosθ

Explanation: (cos theta + i sin theta)<sup>3</sup> = cos 3theta + i sin 3theta. LHS expansion: real part = cos<sup>3theta</sup> - 3cos theta sin<sup>2theta</sup> = cos<sup>3theta</sup> - 3cos theta(1-cos<sup>2theta</sup>) = 4cos<sup>3theta</sup> - 3cos theta.

Argand Plane: z = a + ibReImz = a+iba (real part)b (imaginary part)θ = arg(z)|z| = length of the vector OZ = √(a²+b²); θ = angle OZ makes with the positive real axis

A complex number z = a+ib is plotted as a point (or vector from the origin) on the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical; its modulus |z| is the vector's length, and its argument θ is the angle from the positive real axis.

Concept context

Complex numbers in a+ib form, modulus, argument, polar form, cube roots of unity, and quadratic equations with complex roots

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