Simplify the expression? #cos(270+2alpha)+sin(450+2alpha)#

Question

Simplify:
#cos(270+2alpha)+sin(450+2alpha)#

AND if you know:
#3-4cos^2(3/2pi-alpha)#

There will be:

#3-4sin^2alpha#

OR

#3+4sin^2alpha#

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Answer 1 · 2017-04-09T15:24:54

I got #sin2alpha + cos2alpha#.

I would use the following two formulas:

#cos(A + B) = cosAcosB - sinAsinB#
#sin(A + B) = sinAcosB + sinBcosA#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We have:

#cos(270)cos(2alpha) - sin(270)sin(2alpha) + sin(450)cos(2alpha) + cos(450)sin(2alpha)#

Because #5(90˚) = 450˚# we have #sin450˚ = 1# and #cos450˚ =0#.

#-(-sin2alpha) + 1(cos2alpha) + 0(cos2alpha)#

#sin2alpha + cos2alpha#

I think this is as far as we can simplify.

Hopefully this helps!