prove that COS49= SIN(p+10)

Do you mean find p?  If so then p+10 = 90 - 49, so p = 31°.  Another possibility is in the second quadrant where p + 10 = 180 - 41 so p = 129°  (The 41 comes from the first p+10, i.e. 31+10 = 41), 

Check:

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{49}}^\circ\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{31}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{129}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$