# Going from trig identities to hyperbolic identities

Show how $$\cosh^2z-\sinh^2z=1$$ $$\cosh(z_1+z_2)=\cosh z_1 \cosh z_2 + \sinh z_1 \sinh z_2$$

follows from $$\sin^2z+\cos^2z=1$$ $$\cos(z_1+z_2)=\cos z_1 \cos z_2 - \sin z_1 \sin z_2$$

Approach: For the first identity I was thinking if we set $\sin z=-i\sinh(iz)$ and $\cos z=\cosh(iz)$ in the third identity, so we get $$\cosh^2(iz)-\sinh^2(iz)=1$$

but I don't know how to get rid of the $i$s. I would use a similar approach for the second identity

## 1 Answer

If you consider $z$ as a dummy variable, or let $w=iz$, you already have the identity. Since the $z$ from one identity doesn't necessarily correspond to the $z$ in another.

• we can map any complex number by using z right?, so the identity would satisfy. – TheMathNoob Aug 7 '16 at 6:24
• @TheMathNoob Yes, exactly. So there's no need to eliminate $i$. – lEm Aug 7 '16 at 6:26
• what about the second one?. How would you do it? – TheMathNoob Aug 7 '16 at 6:28
• @TheMathNoob It is exactly the same idea, e.g. the left hand side would be $\cosh (iz_1 + iz_2)$ – lEm Aug 7 '16 at 6:29
• But can I start at cosh and end up at cos? Don't I have to do it the way around? – TheMathNoob Aug 7 '16 at 6:36