1)sin127 cos82- cos127 sin82
2)sin220+ cos220
3) sin4alpha+ sin2alpha• cos2alpha+ cos2alpha=1
4) sin2x-sinx=0
Thanks for the help!!!
1)sin127 cos82- cos127 sin82
Notice that this actually simplifies to sin (127 - 82) = sin (45) = 1 / √2
2)sin^2(20)+ cos^2(20) = 1 [ using the identity sin^2(theta) + cos^2(theta) = 1 ]
3) sin^4alpha+ sin^2alpha• cos^2alpha+ cos^2alpha=1
sin^4alpha + sin^2theta (1 - sin^2theta) + cos^2alpha
sin^4alpha + sin^2alpha - sin^4alpha + cos^2 alpha =
sin^2alpha + cos^2alpha = 1 and this is true
4) sin2x-sinx=0
2sinxcosx - sinx = 0 factor
sinx ( 2cosx - 1) = 0 set each factor to 0 and solve
sin x = 0 at 0 + pi *n where n is an integer
And for the other factor
2cosx - 1 = 0 add 1 to both sides
2cosx = 1 divide both sides by 2
oos x = 1/2
And this happens at pi/3 + 2pi*n and at 5pi/3 + 2pi*n for some integer, n