Evaluate the following expression for sine value?

Question

#sin^2theta -sin^4 theta = 3/16#

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Answer 1 · 2018-08-08T15:31:04

#+- 1/2 +- 1/2 sqrt3#

#3/16 = sin^2theta - sin^4theta = sin^2theta ( 1 - sin^2theta)#

#= sin^2theta cos^2theta = 1/4 sin^2 2theta#

So,

#sin^2 2theta = 3/4 rArr sin 2theta = +-1/2 sqrt 3 = sin ( +_ pi/3)#

#2theta = kpi + (-1)^k ( +-pi/3) rArr theta = k/2pi +(-1)^k(+-pi/6)#

#rArr sin theta = sin ( kpi/2 + (-1)^k ( +-pi/6)), k =0, +-1, +-2, +-3,,,#

# = repetitive { sin ( -2/3pi ) sin ( 1/6pi ) sin ( -1/6pi) sin (2/3pi }#

# = { +-1/2 +-sqrt3/2 }

graph{ y-x^2+x^4+3/16=0[-1 1 -.01 .01]}

x-intercepts are the sine values.