How do you simplify #sin(x-pi/2)#?

2 Answers
Aug 16, 2016

#-cosx#

Explanation:

Using the appropriate #color(blue)"addition formula"#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A-B)=sinAcosB-cosAsinB)color(white)(a/a)|)))#

#rArrsin(x-pi/2)=sinxcos(pi/2)-cosxsin(pi/2)#

now #cos(pi/2)=0" and " sin(pi/2)=1#

#rArrsin(x-pi/2)=sinx(0)-cosx(1)=-cosx#

This is a useful result and worth knowing for future reference.

Dec 15, 2017

#-cosx #

Explanation:

The answer given prior is a perfectly valid explanation, but here is another:

We must consider our knowledge of transformations:

#f(x-alpha) # is a translation of #f(x) # by # (alpha,0) #

So hence:

#sin(x- pi/2)# is just #sinx# translated by #(pi/2,0)#

We see that #sinx#: graph{sinx [-4.006, 4.006, -2.003, 2.003]}

#therefore sin(x-pi/2) #: graph{sin(x- pi/2) [-4.006, 4.006, -2.003, 2.003]}

By observing this new graph, we see that this is just #cosx# reflected in the #x# axis

Hence just:

#=> -cosx #