Evaluating $\sin(5-8i)$

Question

We know that $$\sin(a\pm b)=\cos(b)\sin(a)\pm \cos(a)\sin(b)$$

We also know that $$\sinh(x)=-i\,\sin(ix) \quad\text{and}\quad\cosh(x)=\cos(ix)$$

Now, suppose I want to compute $\sin(5-8i)$.

Then, $$\sin(5-8i)=\cos(8i)\sin(5)-\sin(8i)\cos(5) \tag{1}$$

Now we substitute trigonometric definitions of hyperbolic functions in (1). Then, we get,

$$\begin{align} \sin(5-8i) &=\cosh(8)\sin(5)-\cos(5)(-i\,\sinh(8)) \\ &=\cosh(8)\sin(5)+i\,\cos(5)\sinh(8) \end{align}\tag{2}$$

My hp 50g calculator gives answer to $\sin{(5-8i)}$ in radian mode, $$(-1429.2566486,-422.79248111)$$ But R.H.S. of $(2)$ is $$(-1429.2566486,422.79248111)$$

Where am I wrong in this computation?

Any member knowing the correct answer to this question may reply with correct answer.

Answer 1

Starting with

$$\sin(a- b)=\cos(b)\sin(a)- \cos(a)\sin(b)\tag{1}$$ $$\sinh(x)=-i\,\sin(ix)\tag{2}$$ $$\cosh(x)=\cos(ix)\tag{3}$$

we see that $(2)$ can be rewritten as

$$i\sinh(x)=i\big(-i\,\sin(ix)\big)\implies i\sinh(x)=\sin(ix)\tag{4}$$

therefore

\begin{align}\sin(5-8i)&=\cos(8i)\sin(5)-\cos(5)\sin(8i)\qquad\text{from (1)}\\&= \cosh(8)\sin(5)-\cos(5)\sin(8i)\qquad\text{from (3)}\\&= \cosh(8)\sin(5)-i\cos(5)\sinh(8)\qquad\text{from (4)} \end{align}

which matches the result provided by your calculator.

Answer 2

We have $$\sin(5-8i)=\sin \left( 5 \right) \cosh \left( 8 \right) -i\cos \left( 5 \right) \sinh \left( 8 \right) $$

Answer 3

Hint: $$ \sin z = \frac{e^{iz}-e^{-iz}}{2i} $$