what are the values of cos(theta)=sin(thesta)sin2(theta)

Answer 1

1. cosT=sinTsin2T
2. cosT=sinT2sinTcosT
3. 1=2sin^2(T)
4. 1/2 =sin^2(T)
5. sinT=+/-sqr(1/2)=+/-1/sqr2=+/-sqr2/2
6. T=+/-pi/4 or +/-45 degrees +2npi where n = any integer

check the answers

cos45 = sqr2/2

sin45sin2(45)= sin45sin90 = sqr2/2(1) = sqr2/2

cos-45 = sqr2/2

sin-45sin2(-45)= sin-45sin-90= -sqr2/2)(-1) = sqr2/2

another solution

cosT=0

T = 90 degrees, 270 degrees or pi/2, 3pi/2 +2npi where n= any integer

cos90 = 0

sin90sin2(90) = sin90sin180=1(0) = 0

full soluitions

T=45,-45,0,270, each +360ni where n= any integer

or

T=+/-pi/2, pi,3pi/2 each with +2npi where n=any integer

UNLESS you meant cosT=sinTsin^2(T)

then

cosT=sin^3(T)

+/-sqr(1-sin^2(T))=sin^3(T)

1-sin^2(T)= sin^6(T)

let x=sin^2(T)

x^3 +x -1 =0

x=.682328

sinT=+/-sqr.682328=about +/-.826

T= sin^-1(+/-.826)

T= +/-55.69 +360n where n=any integer

= about +/-.972 radians

= about +/-1 radian +2pn

=about +/-1+2npi radians