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

1 Answer
Jun 2, 2016

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)#

Tony B

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

'.................................................

#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")#

Tony B