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)

Telangana Board Solutions for Exercise 1: Exercise 2(a)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana 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 Telangana Board Solutions for Exercise 1: Exercise 2(a) with Hints & Solutions

MEDIUM

12th Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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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