If α , β are the roots of the equation x 2 - 2 x + 4 = 0 , then for any n ∈ N , show that α n + β n = 2 n + 1 cos n π 3 .

Given quadratic equation: x 2 - 2 x + 4 = 0 and to prove: α n + β n = 2 n + 1 cos n π 3 ⇒ x = 2 ± 4 - 16 2 = 2 ± 2 i 3 2 = 1 ± i 3 ∵ x = - b ± b 2 - 4 a c 2 a , when a x 2 + b x + c = 0 Given that, α , β are the roots of the given quadratic equation i.e., α = 1 + i 3 = 2 1 2 + i 3 2 = 2 cos π 3 + i sin π 3 and β = 1 - i 3 = 2 1 2 - i 3 2 = 2 cos π 3 - i sin π 3 Now, from LHS, we have, α n + β n = 2 cos π 3 + i sin π 3 n + 2 cos π 3 - i sin π 3 n = 2 n cos n π 3 + i sin n π 3 + cos n π 3 - i sin n π 3 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = 2 n 2 cos n π 3 = 2 n + 1 cos n π 3 = RHS Hence, proved.

If n is an integer and z = cis θ , θ ≠ ( 2 n + 1 ) π 2 , then show that z 2 n - 1 z 2 n + 1 = i tan n θ .

Given that : z = cis θ = cos θ + i sin θ , θ ≠ ( 2 n + 1 ) π 2 , where n is an integer. To prove: z 2 n - 1 z 2 n + 1 = i tan n θ From, LHS, we have, = z 2 n - 1 z 2 n + 1 = ( cos θ + i sin θ ) 2 n - 1 ( cos θ + i sin θ ) 2 n + 1 = cos 2 n θ + i sin 2 n θ - 1 cos 2 n θ + i sin 2 n θ + 1 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = - ( 1 - cos 2 n θ ) + i sin 2 n θ ( 1 + cos 2 n θ ) + i sin 2 n θ = i 2 2 sin 2 n θ + 2 i sin n θ · cos n θ 2 cos 2 n θ + 2 i sin n θ · cos n θ ∵ - 1 = i 2 , 1 + cos 2 x = 2 cos 2 x , 1 - cos 2 x = 2 sin 2 x & sin 2 x = 2 sin x cos x = 2 i sin n θ { cos n θ + i sin n θ } 2 cos n θ { cos n θ + i sin n θ } = i tan n θ = RHS Hence, proved.

If ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + … . + a n x n , then show that a 0 - a 2 + a 4 - a 6 + … = 2 n 2 cos n π 4 .

Given: ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + … + a n x n and To prove: a 0 - a 2 + a 4 - a 6 + … = 2 n 2 cos n π 4 Put x = i , then we get: ( 1 + i ) n = a 0 + a 1 i + a 2 i 2 + … + a n i n ⇒ 2 1 2 + 1 2 i n = a 0 + a 1 i - a 2 + … + a n i n ⇒ 2 1 2 + 1 2 i n = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ⇒ 2 cos π 4 + i sin π 4 n = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ⇒ 2 n 2 cos n π 4 + i sin n π 4 = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z Now, Equating Real parts both sides, we get: a 0 - a 2 + a 4 - a 6 + … = 2 n 2 cos n π 4 Hence, proved.

If ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + … . + a n x n , then show that a 1 - a 3 + a 5 + … = 2 n 2 sin n π 4 .

Given: ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + … + a n x n and To prove: a 0 - a 2 + a 4 - a 6 + … = 2 n 2 cos n π 4 Put x = i , then we get: ( 1 + i ) n = a 0 + a 1 i + a 2 i 2 + … + a n i n ⇒ 2 1 2 + 1 2 i n = a 0 + a 1 i - a 2 + … + a n i n ⇒ 2 1 2 + 1 2 i n = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ⇒ 2 cos π 4 + i sin π 4 n = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ⇒ 2 n 2 cos n π 4 + i sin n π 4 = a 0 - a 2 + a 4 - a 6 … + i a 1 - a 3 + a 5 - a 7 … ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z Now, Equating Imaginary parts on both sides, we get: a 1 - a 3 + a 5 + … = 2 n 2 sin n π 4 Hence, proved.

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 2 - De Moivre's Theorem>Exercise 2(a)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 2(a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 2(a)

Attempt the free practice questions from Exercise 1: Exercise 2(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 1: Exercise 2(a) with Hints & Solutions

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q I

If ${(\frac{\sqrt{3}}{2} + \frac{i}{2})}^{5} - {(\frac{\sqrt{3}}{2} - \frac{i}{2})}^{5} = a + i b$ then find the value of $b$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $α, β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$ , then for any $n \in N$ , show that $α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $\cos α + \cos β + \cos γ = 0 = \sin α + \sin β + \sin γ$ , then show that $\cos 3 α + \cos 3 β + \cos 3 γ = 3 \cos (α + β + γ)$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $\cos α + \cos β + \cos γ = 0 = \sin α + \sin β + \sin γ$ , then show that $\sin 3 α + \sin 3 β + \sin 3 γ = 3 \sin (α + β + γ)$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $\cos α + \cos β + \cos γ = 0 = \sin α + \sin β + \sin γ$ , then show that $\cos (α + β) + \cos (β + γ) + \cos (γ + α) = 0$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $n$ is an integer and $z = cis θ, (θ \neq (2 n + 1) \frac{π}{2})$ , then show that $\frac{z^{2 n} - 1}{z^{2 n} + 1} = i \tan n θ$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{n} x^{n}$ , then show that $a_{0} - a_{2} + a_{4} - a_{6} + \dots = 2^{\frac{n}{2}} \cos \frac{n π}{4}$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(a)>Q II

If $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{n} x^{n}$ , then show that $a_{1} - a_{3} + a_{5} + \dots = 2^{\frac{n}{2}} \sin \frac{n π}{4}$ .

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