How do you graph #y = cos(90-A)# as sinA?

1 Answer
Lovecraft
Sep 17, 2015

Just graphing it. #cos(90 - A) = sin(A)#

Explanation:

There are many ways to go around and show this, the easiest is simply plugging the 5 main values and show that it matches (even though this isn't a proof it's enough to understand)

#cos(90 - 0) = cos(90) = 0 = sin(0)#
#cos(90 - 30) = cos(60) = 1/2 = sin(30)#
#cos(90 - 45) = cos(45) = sqrt2/2 = sin(45)#
#cos(90 - 60) = cos(30) = sqrt3/2 = sin(60)#
#cos(90 - 90) = cos(0) = 1 = sin(90)#
#cos(90 - 180) = cos(-90) = cos(90) = 0 = sin(180)#

We can proof this though, by using the property
#cos(a - b) = cos(a)cos(b) + sin(a)sin(b)#

So

#cos(90 - x) = cos(90)cos(x) + sin(90)sin(x)#

As we've seen on the list above, #cos(90) = 0# and #sin(90) = 1#, thus:

#cos(90 - x) = 0cos(x) + 1sin(x)#
#cos(90 - x) = sin(x)#

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