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Question: If \(\cos(x + iy) = A + iB\), then A equals...

If cos⁡(x+iy)=A+iB\cos(x + iy) = A + iB, then A equals

A

cos⁡xcosh⁡y\cos x\cosh y

B

sin⁡xsin⁡hy\sin x{\sin h}y

C

−sin⁡xsinh⁡y- \sin x\sinh y

D

cos⁡xsin⁡hy\cos x{\sin h}y

Answer

cos⁡xcosh⁡y\cos x\cosh y

Explanation

Solution

cos⁡(x+iy)=A+iB\mathbf{\cos}\mathbf{(}\mathbf{x + iy) = A + iB}

⇒cos⁡xcos⁡(iy)−sin⁡xsin⁡(iy)=A+iB\mathbf{\cos}\mathbf{x}\mathbf{\cos}\mathbf{(}\mathbf{iy)}\mathbf{-}\mathbf{\sin}\mathbf{x}\mathbf{\sin}\mathbf{(}\mathbf{iy) = A + iB}

⇒ cos⁡xcosh⁡y−isin⁡xsinh⁡y=A+iB∴\cos x\cosh y - i\sin x\sinh y = A + iB\therefore A=cos⁡xcos⁡hyA = \cos x{\cos h}y