# trig putting in for rcos(x-a)

• January 16th 2009, 07:52 AM
rpatel
trig putting in for rcos(x-a)
Hello

I need help with some this type of trig question

can you someone please go throught the process of solving it so thats its in the form that is required. I am unsure on how to do these type of questions

thanks
• January 16th 2009, 08:28 AM
running-gag
Hi

$a cos x + b sin x = \sqrt{a^2 + b^2} \:\left( \frac{a}{\sqrt{a^2 + b^2}} cos x + \frac{b}{\sqrt{a^2 + b^2}} sin x \right)$

$\frac{a}{\sqrt{a^2 + b^2}}$ and $\frac{b}{\sqrt{a^2 + b^2}}$ can be seen respectively as the cosines and the sinus of the same angle $\theta$ because the sum of their square is equal to 1

Therefore

$a cos x + b sin x = \sqrt{a^2 + b^2} \:\left( cos \theta cos x + sin \theta sin x \right) = \sqrt{a^2 + b^2} \:cos(x-\theta)$