Show that β - α= π/ 2; then sin(x+α) + cos(x+β) = 0?
I really don't get how to approach this, this is what I have so far for the first part: β= π/ 2 + α and α= -π/ 2 + β, but I don't think it's right.
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cosβ = cos(π/2 + α) = cosπ/2 cosα - sinπ/2 sinα = -sinα,
because cosπ/2=0 and sinπ/2=1.
sinβ = sin(π/2 + α) = sinπ/2 cosα + cosπ/2 sinα = cosα,
because cosπ/2=0 and sinπ/2=1.
cos(x+β) = cosxcosβ - sinxsinβ = cosx cos(π/ 2 + α) - sinx sin(π/ 2 + α) =
cosx (-sinα) - sinx cosα = -(sinx cosα + cosx sinα) = - sin(x+α).
So sin(x+α) + cos(x+β) = sin(x+α) - sin(x+α) = 0.