$$e^{\theta(s\times)} = I + \sin\theta(s\times) + (1 − \cos \theta)(s\times)^2$$
I have to prove the above formula and am not sure where to start, may someone please help me! The full question is posted on this link
https://i.stack.imgur.com/UNGmM.jpg
$(s\times)$ is the skew operator and it is defined below, it pretty much is the cross product
$$s\times \triangleq \left[\begin{array}{ccc} 0&-c_s&b_s\\c_s&0&-a_s\\-b_s&a_s&0 \end{array}\right]$$
The formula is true only when $s$ is a unit vector. The key is that $(s\times)^3=-(s\times)$. For convenience, write $K=(s\times)$. Then \begin{align*} e^{\theta K} &=I+\theta K+\frac1{2!}\theta^2K^2+\frac1{3!}\theta^3K^3+\frac1{4!}\theta^4K^4 +\frac1{5!}\theta^5K^5+\frac1{6!}\theta^6K^6+\ldots\\ &=I+\theta K+\frac1{2!}\theta^2K^2-\frac1{3!}\theta^3K-\frac1{4!}\theta^4K^2 +\frac1{5!}\theta^5K+\frac1{6!}\theta^6K^2+\ldots\\ &=I+\left(\theta-\frac1{3!}\theta^3+\frac1{5!}\theta^5-\ldots\right)K +\left(\frac1{2!}\theta^2-\frac1{4!}\theta^4 +\frac1{6!}\theta^6+\ldots\right)K^2 \end{align*} and the result follows.