Question

How do you prove `cos(x-(pi/2))=sin x`?

Answer 1

See explanation.

Explanation:

There is a group of Trig Identities that contain:

`cos(A-B)=cos(A)cos(B)+sin(A)sin(B)`

For your question this translates to:

`cos(x-pi/2)=cos(x)cos(pi/2)+sin(x)sin(pi/2)`

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`color(brown)("Condition 1")`
`cos(pi/2)->cos(90^o) = ("adjacent")/("hypotenuse")=0/h=0`

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`color(brown)("Condition 2")`

`sin(pi/2)->sin(90^o)=("opposite")/("hypotenuse")=h/o`

but for this condition `h=o`

`=>sin(pi/2)=1`
'......................................................

Thus we have:

`cos(x-pi/2)=[cos(x)xx0]+[sin(x)xx1]`

`color(green)(=> cos(x-pi/2)= sin(x)" QED")`