How do you prove $cos(x-pi/6) = 1/2([sqrt3] cosx + sin x)$ ?

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Answer 1 · 2016-04-16T23:30:05

See below

LHS =left hand side, RHS =right hand side

Use formula $cos(A-B)=cosA cosB + sinA sin B$

LHS $=cos x cos (pi/6) + sinx sin (pi/6)$

$=cos x xx sqrt 3/2 + sinx xx 1/2$

$=1/2 (sqrt 3 cosx + sinx)$

$= RHS$

How do you prove cos(x-pi/6) = 1/2([sqrt3] cosx + sin x)cos(x-pi/6) = 1/2([sqrt3] cosx + sin x)?