Use the subtraction theorem of cosine.** cos(x-5π/4) = cosx·cos(5π/4) + sinx·sin(5π/4) Here, cos(5π/4) = cos(π + π/4) = -cos(π/4), sin(5π/4) = sin(π + π/4) = -sin(π/4) Thus, cos(x - 5π/4) = -cosx·cos(π/4) - sinx·sin(π/4) = -1 x {cosx·cos(π/4) + sinx·sin(π/4)} The reverse of subtraction theorem of cosine is also true. So, {cosx·cos(π/4) + sinx·sin(π/4)} = cos(x - π/4) Therefore, cos(x - 5π/4) = -cos(x - π/4) ··· Eq.1
While, cos(π/4) = cos( π/2 - π/4) = sin(π/4) So, {cosx·cos(π/4) + sinx·sin(π/4)} = cosx·sin(π/4) +sinx·cos(π/4) = sinx·cos(π/4) + cosx·sin(π/4) = sin(x + π/4) Therefore, cos(x-5π/4) = -sin(x+π/4) ··· Eq.2
From Eq.1 andEq.2 we have, cos(x - 5π/4) = -cos(x - π/4) = -sin(x+π/4)
**: Use the symmetry of trigonometric functions for certain angles, instead of the compound formula, for example cos(-θ) = cosθ, cos (π-θ) = -cosθ, or cos(π/2-θ) = sinθ.
(5π/4) = π+π/4 So, cos(x-5π/4) = cos(5π/4-x) = cos{π-(x-π/4)} = -cos(x-π/4) ···Eq.3
While, π/4 = π/2- π/4 So, cos(x-π/4) = cos(π/4-x) = cos{π/2-(x+π/4)} = sin(x+π/4) ··· Eq.4
From Eq.3 and Eq.4 we have, cos(x-5π/4) = -cos(x-π/4) = -sin(x+π/4)