Find R and a such that 7*cos(x)+3*sin(x)=Rcos(x-a)

Use the cosine trig identity, cos(a-b)=cos(a)cos(b)+sin(a)sin(b) ,to write Rcos(x-a) as R(cos(x)cos(a)+sin(x)sin(a)).Now we can equate the coefficients of the sines and cosins on either side of the equation givingRcos(a)=3 , and Rsin(a)=7Now knowing Pythagoras's theorem we can square these equations and sum themR^2 * cos(a)^2 + R^2 * sin(a)^2 = 3^2 + 7^2 (factor out the R^2)= R^2(cos(a)^2+sin(a)^2)=3^2+7^2 (by Pythagoras cos(a)^2+sin(a)^2=1)so we have R^2=3^2+7^2 so R=sqrt(7^2+3^2).Now we need to find a.We know that tan(a)=sin(a)/cos(a)so using this we can writeRsin(a)/Rcos(a)=tan(a)=3/7 hence a=arctan(3/7) = 0.404...