Prove that cos(x + pi/6) + sin(x + pi/3) = (sqrt 3)cos x?

Question

Prove that cos(x + pi/6) + sin(x + pi/3) = (sqrt 3)cos x

Thanks for any help!

Dan

Answer 1

the key to this problem is recalling the trig formulas of:

cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

and

sin(a+b) = sin(a)cos(b)+cos(a)sin(b)

this gives:

cos(x)cos(pi/6)-sin(x)sin(pi/6) + sin(x)cos(pi/3)+cos(x)sin(pi/3) = (sqrt 3)cos(x)

not use right triangles or recall the unit circle to get that:

cos(pi/6)=sqrt(3)/2
sin(pi/6)=1/2
cos(pi/3)=1/2
sin(pi/3)=sqrt(3)/2

this gives:

cos(x)sqrt(3)/2 - sin(x)/2 + sin(x)/2 + cos(x)sqrt(3)/2 = sqrt(3)cos(x)

the two middle terms drop out and the other two can be added together to get:

2cos(x)sqrt(3)/2 = sqrt(3)cos(x)

cancel the 2's on the right and you've got your answer.

Answer 2

cos(x+pi/6)+sin(x+pi/3)=cosxcos pi/6- sinx sinpi/3+sinxcospi/3+cosx sinpi/3
=sqrt(3)/2 cosx-sqrt(3)/2 sinx+1/2 sinx+sqrt(3)/2 cosx
=sqrt(3)cosx+sinx[sqrt(3-1/2)].ans
=

Answer 3

cos x cos π/6 - sin x sin π/6
+
sin x cos π/3 + cos x sin π/3

(√3 / 2) cos x - (1/2) sin x
+
(1/2) sin x + (√3/2) cos x

√3 cos x