Can anyone solve this?

sin (90-x) degrees = cos x degrees? What steps are there to get the answer? Thanks


rr

There's a rule that states:

sin(x-y) = sin(x)cos(y) - cos(x)sin(y)

So:

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

Streetwolf,

Have you taken the DAT yet? Also, I have another Trig. problem. I have not had math in a while so that is why I am trying to learn a little trig before my test Saturday on the 9th.
Problem:
Cos-pi/6 (I cannot put the pi sign here from my computer). Thanks in advance for any suggestions on learn trig.

rr

Yes I took the DAT last June.

cos(pi/6) = sqrt(3) / 2

It would help to learn the basic trig function values:

sin(0) = 0
sin(pi/6) = 1/2
sin(pi/4) = sqrt(2) / 2
sin(pi/3) = sqrt(3) / 2
sin(pi/2) = 1
sin(pi) = 0
sin(3pi/2) = -1
sin(2pi) = 0

cos(0) = 1
cos(pi/6) = sqrt(3) / 2
cos(pi/4) = sqrt(2) / 2
cos(pi/3) = 1/2
cos(pi/2) = 0
cos(pi) = -1
cos(3pi/2) = 0
cos(2pi) = 1

Plus the identities:

cos(90 - x) = sin(x)
sin(90 - x) = cos(x)
sin(x + y) = sin(x)cos(y) + sin(y)cos(x)
sin(x - y) = sin(x)cos(y) - sin(y)cos(x)
sin(2x) = sin(x + x) = 2sin(x)cos(x)
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
cos(2x) = cos(x + x) = cos^2(x) - sin^2(x)
tan(x) = sin(x)/cos(x)
sec(x) = 1/cos(x)
csc(x) = 1/sin(x)
cot(x) = 1/tan(x) = cos(x)/sin(x)

sin^2(x) + cos^2(x) = 1

Divide the above by sin^2(x) to get:

1 + cot^2(x) = csc^2(x)

Divide it by cos^2(x) instead to get:

tan^2(x) + 1 = sec^2(x)

And finally:

cos^2(x) = 1/2 * (1 + cos(2x))
sin^2(x) = 1/2 * (1 - cos(2x))

You should double check a lot of those more complex identities to see how they are derived from the basic identities.

Good luck!