Express 3cos(theta) + 5sin(theta) in the form Rcos(theta - alpha) where R and alpha are constants, R>0 and 0<alpha<90. Give the exact value of R and the value of alpha to 2dp.

Write out identity:

Rcos(theta - alpha) = Rcos(theta)cos(alpha) + Rsin(theta)sin(alpha) from formula booklet

Write out in form of question so it's easier to compare:

3cos(theta) + 5sin(theta) = R[cos(alpha)cos(theta) + sin(alpha)sin(theta)]

By comparing either side, you can see that Rcos(alpha) = 3 (equation 1) and Rsin(alpha) = 5 (equation 2)

Another identy you should know is sin^2(alpha) + cos^2(alpha) = 1

Therefore:

R^2cos^2(alpha) + R^2sin^2(alpha) = 3^2 + 5^2

Factoring out R^2:

R^2[cos^2(alpha) + sin^2(alpha)] = 9 + 25

Using identity sin^2(alpha) + cos^2(alpha) = 1:

R^2 = 34

R = root(34)

By dividing equation 2 by equation 1:

Rsin(alpha) / Rcos(alpha) = 5/3

Cancelling R, and sin/cos = tan so:

tan(alpha) = 5/3

alpha = tan^-1(5/3)

alpha = 59.04 degrees (2dp)

Therefore 3cos(alpha) + 5sin(alpha) = root(34)cos(theta - 59.04)