Question

m,n are integer and x=cos alpha + i sin alpha , y= cos beta + i sin beta then prove that x^m y^n + 1/x^m y^n = cos (m alpha +n beta) and x^m y^n - 1/x^m y^n = 2 i sin (m alpha + n beta)

Answer 1

Answer:

$\displaystyle{\sf\:i\:)\:\boxed{\red{\sf\:x^m\:y^n\:+\:\dfrac{1}{x^m\:y^n}\:=\:2\sf\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}}}$

$\displaystyle{\sf\:ii\:)\:\boxed{\blue{\sf\:x^m\:y^n\:-\:\dfrac{1}{x^m\:y^n}\:=\:2\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:x\:=\:\cos\:\alpha\:+\:i\:\sin\:\alpha}$

$\displaystyle{\sf\:y\:=\:\cos\:\beta\:+\:i\:\sin\:\beta}$

We have to prove that,

$\displaystyle{\sf\:i\:)\:x^m\:y^n\:+\:\dfrac{1}{x^m\:y^n}\:=\:\textbf{\textsf2}\sf\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\sf\:ii\:)\:x^m\:y^n\:-\:\dfrac{1}{x^m\:y^n}\:=\:2\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

Now,

$\displaystyle{\sf\:x\:=\:\cos\:\alpha\:+\:i\:\sin\:\alpha}$

By Euler's formula,

$\displaystyle{\pink{\sf\:e^{i\:x}\:=\:\cos\:x\:+\:i\:\sin\:x}}$

$\displaystyle{\implies\sf\:x\:=\:e^{i\:\alpha}}$

$\displaystyle{\implies\sf\:x^m\:=\:(\:e^{i\:\alpha}\:)^m}$

$\displaystyle{\implies\sf\:x^m\:=\:e^{i\:\alpha\:m}\:\qquad\cdots\:(\:1\:)}$

Now,

$\displaystyle{\sf\:y\:=\:\cos\:\beta\:+\:i\:\sin\:\beta}$

By Euler's formula,

$\displaystyle{\implies\sf\:y\:=\:e^{i\:\beta}}$

$\displaystyle{\implies\sf\:y^n\:=\:(\:e^{i\:\beta}\:)^n}$

$\displaystyle{\implies\sf\:y^n\:=\:e^{i\:\beta\:n}\:\qquad\cdots\:(\:2\:)}$

Now, we have to prove that,

$\displaystyle{\sf\:x^m\:y^n\:+\:\dfrac{1}{x^m\:y^n}\:=\:2\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:x^m\:y^n\:+\:\dfrac{1}{x^m\:y^n}}$

$\displaystyle{\implies\sf\:LHS\:=\:x^m\:y^n\:+\:(\:x^m\:y^n\:)^{-\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:(\:e^{i\:\alpha\:m}\:)\:(\:e^{i\:\beta\:n}\:)\:+\:[\:(\:e^{i\:\alpha\:m}\:)\:(\:e^{i\:\beta\:n}\:)\:]^{-\:1}\:\quad\cdots[\:From\:(\:1\:)\:\&\:(\:2\:)\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:\alpha\:m\:+\:i\:\beta\:n}\:+\:(\:e^{i\:\alpha\:m\:+\:i\:\beta\:n}\:)^{-\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)}\:+\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)^{-\:1}}}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)}\:+\:e^{-\:i\:(\:m\:\alpha\:+\:n\:\beta\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:-\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:2\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\sf\:RHS\:=\:2\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\therefore{\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!

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Now,

We have to prove that,

$\displaystyle{\sf\:ii\:)\:x^m\:y^n\:-\:\dfrac{1}{x^m\:y^n} \:=\:2\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:x^m\:y^n\:-\:\dfrac{1}{x^m\:y^n}}$

$\displaystyle{\implies\sf\:LHS\:=\:x^m\:y^n\:-\:(\:x^m\:y^n\:)^{-\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:(\:e^{i\:\alpha\:m}\:)\:(\:e^{i\:\beta\:n}\:)\:-\:[\:(\:e^{i\:\alpha\:m}\:)\:(\:e^{i\:\beta\:n}\:)\:]^{-\:1}\:\quad\cdots[\:From\:(\:1\:)\:\&\:(\:2\:)\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:\alpha\:m\:+\:i\:\beta\:n}\:-\:(\:e^{i\:\alpha\:m\:+\:i\:\beta\:n}\:)^{-\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)}\:-\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)^{-\:1}}}$

$\displaystyle{\implies\sf\:LHS\:=\:e^{i\:(\:m\:\alpha\:+\:n\:\beta\:)}\:-\:e^{-\:i\:(\:m\:\alpha\:+\:n\:\beta\:)}}$

$\displaystyle{\implies}\sf\:LHS\:=\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)\:-\:[\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:-\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)\:]$

$\displaystyle{\implies\sf\:LHS\:=\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)\:-\:\cos\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)\:+\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:2\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\displaystyle{\sf\:RHS\:=\:2\:i\:\sin\:(\:m\:\alpha\:+\:n\:\beta\:)}$

$\therefore{\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!

Answer 2

Step-by-step explanation:

given :

• m,n are integer and x=cos alpha + i sin alpha , y= cos beta + i sin beta then prove that x^m y^n + 1/x^m y^n = cos (m alpha +n beta) and x^m y^n - 1/x^m y^n = 2 i sin (m alpha + n beta)

to find :

• m y^n = cos (m alpha +n beta) and x^m y^n - 1/x^m y^n = 2 i sin (m alpha + n beta)

solution :

• check the attached file there is your answer