Express each of the following in the form Rsin(x+a), where r>0 and 0< a <2pi

Question

(a) 5cosx + 12sinx
(b) 12cosx + 5sinx

Answer 1

I will do the first, and you try the second:

let Rsin(x+a) 5cosx + 12sinx

Rsin(x+a) = R(sinxcosa + cosxsina)
= Rsinxcosa + Rcosxsina

so

Rsinxcosa + Rcosxsina = 5cosx + 12sinx
if this is an identity it must be true for all values of x

Pick any two values of x which you can easily find

let x = 0
Rsin0cosa + Rcos0sina = 5cos0 + 12 sin0
but sin0 = 0 , and cos0 = 1
Rsina = 5
sina = 5/R

let x = 90° or π/2
Rsin90cosa + Rcosasin90 = 5cos90 + 12sin90
sin90 = 1 , and cos90 = 0
Rcosa = 12
cosa = 12/R

but sin^2 a + cos^2 a = 1
25/R^2 + 144/R^2 = 1
R^2 = 169
R = 13

also sina/cosa = (5/R)/(12/R)
tana = 5/12
a = 22.6199°

So 5cosx + 12sinx = 13sin(x + 22.6199°)
or if you want radians, set your calculator to radians when you find the tangent inverse
5sinx + 12cosx = 13sin(x + .3948) in radians

check for any angle
e.g. x = 53°
5cos53 + 12sin53 = 12.5927..
13sin(53+22.6199) = 13sin 75.6199 = 12.5927..

not bad

Answer 2

I noticed a small typo
2nd line clearly should have said

let Rsin(x+a) = 5cosx + 12sinx

Answer 3

12cosx+5sinx=13sin(x+67.38)