# Evaluating $\sin(5-8i)$

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?

• $\sinh(x) = -i \sin(ix)$, so (multiplying both sides by $i$) we have $$i\sinh(x)=\sin(ix)$$ (not "$-i\sinh(x)$").
– Blue
Commented Oct 26, 2019 at 17:14
• @Blue, Thanks, for finding out where i was wrong. Commented Oct 26, 2019 at 17:23
• Sign errors are an ever-present danger, but they can be extra-sneaky when $i$ is involved. Stay vigilant! :)
– Blue
Commented Oct 26, 2019 at 17:26

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.

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

• Graubner, But $\sin{(8*i)}=-i*\sinh{(8)}$. I just substituted it in $\sin{(a \pm b)}=\sin{(a)}*\cos{(b)} \pm \cos{(a)}*\sin{(b)}$ Commented Oct 26, 2019 at 17:14
• Yeah, my name is Sonnhard. You must use the formula $$\sin(\alpha-\beta)$$ Commented Oct 26, 2019 at 17:16

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