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I started with

$cos(q) = cos(a+ib+jc+kd)$

$cos(q) = cos(a)cos(ib+jc+kd)-sin(a)sin(ib+jc+kd)$

$cos(q) = cos(a)(cos(ib)(cos(jc)cos(kd)-sin(jc)sin(kd))-sin(ib)(sin(jc)cos(kd)+cos(jc)sin(kd)))-sin(a)(sin(ib)(cos(jc)cos(kd)-sin(jc)sin(kd))+cos(ib)(sin(jc)cos(kd)+cos(jc)sin(kd)))$

$cos(q) = cos(a)(cosh(b)(jcosh(c)kcosh(d)-jsinh(c)ksinh(d))-isinh(b)(jsinh(c)kcosh(d)+jcosh(c)ksinh(d)))-sin(a)(isinh(b)(jcosh(c)kcosh(d)-jsinh(c)ksinh(d))+icosh(b)(jsinh(c)kcosh(d)+jcosh(c)ksinh(d)))$

then with the help of $i^2=j^2=k^2=ijk=-1$

I got to

$cos(q) = cos(a)(cosh(b)(icosh(c)cosh(d)-isinh(c)sinh(d))-isinh(b)(isinh(c)cosh(d)+icosh(c)sinh(d)))-sin(a)(isinh(b)(icosh(c)cosh(d)-isinh(c)sinh(d))+icosh(b)(-isinh(c)cosh(d)+icosh(c)sinh(d)))$

and finally to

$cos(q) = icos(a)(cosh(b)cosh(c-d)+sinh(b)sinh(c+d))+sin(a)cosh(c-d)(sinh(b)-cosh(b))$

I'd like to know if it is correct, if it is a legitimate function, since I could't find a quaternionic trigonometric function anywhere on the internet. I found it strange since j and k don't appear in the end product.

P.S. I also tried expanding it with Taylor series but it found it cumbersome.

Nilson Fernandes
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1 Answers1

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Given $q=a+bi+cj+dk=(a,\vec{v})$, we define:

$$ \cos(q)=\left(\cos(a)\cosh(|\vec{v}|), -\sin(a)\sinh(|\vec{v}|)\frac{\vec{v}}{|\vec{v}|}\right) $$

where $|\vec{v}|=\sqrt{b^2+c^2+d^2}$.

Example: $$q=1+i+2j+k$$

$$\cos(q)=\left(\cos(1)\cosh(\sqrt6), \frac{-\sin(1)\sinh(\sqrt6)}{\sqrt6}\space(i+2j+k)\right)$$

Tommaso
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