Question

Prove that all values of theta: `|(sintheta, costheta, sin2theta),(sin(theta+(2pi)/3), cos(theta(2pi)/3), sin (2theta+(4pi)/3)),(sin (theta- (2pi)/3), cos (theta- (2pi)/3), sin (2theta- (4pi)/3))|=0`

Answer 1

`"Let"Delta = |{:(" sin"theta," cos"theta, " sin"2theta),("sin"(theta+(2pi)/(3)), "cos"(theta+(2pi)/(3)), "sin"(2theta+(4pi)/(3))),("sin"(theta-(2pi)/(3)),"cos"(theta-(2pi)/(3)), "sin"(2theta-(4pi)/(3))):}|`
Applying `R_(2) to R_(2) + R_(3)`
`= |{:(" sin"theta," cos"theta, " sin"2theta),("sin"(theta+(2pi)/(3)), "cos"(theta+(2pi)/(3)), "sin"(2theta+(4pi)/(3))),(+"sin"(theta-(2pi)/(3)), + "cos" (theta-(2pi)/(3)), +"sin" (2theta-(4pi)/(3))),("sin"(theta-(2pi)/(3)),"cos"(theta-(2pi)/(3)), "sin"(2theta-(4pi)/(3))):}|`
Now, `"sin" (theta + (2pi)/(3)) + "sin" (theta - (2pi)/(3))`
`= 2 "sin" ((theta + (2pi)/(3) + theta - (2pi)/(3))/(2)) "cos" ((theta + (2pi)/(3) - theta + (2pi)/(3))/(2))`
` = 2 "sin" theta "cos" (2pi)/(3) = 2 "sin"theta "cos" (pi - (pi)/(3))`
` = -2 "sin" theta "cos" (pi)/(3) = -"sin" theta`
`"and cos" (theta + (2pi)/(3) + "cos" (theta - (2pi)/(3))`
`= 2 "cos" ((theta + (2pi)/(3) + theta - (2pi)/(3))/(2)) "cos" ((theta + (2pi)/(3) - theta + (2pi)/(3))/(2))`
` =2 "cos" theta "cos" ((2pi)/(3) = 2 "cos" theta (-(1)/(2)) =-"cos" theta`
`"and sin" (2theta + (4pi)/(3)) + "sin" (2theta-(4pi)/(3))`
`= 2 "sin" ((2theta + (4pi)/(3) + 2theta - (4pi)/(3))/(2)) "cos" ((2theta + (4pi)/(3) - 2theta + (4pi)/(3))/(2))`
` = 2"sin" 2theta "cos"(4pi)/(3) = 2"sin"2theta "cos" (pi + (pi)/(3))`
` = -2"sin" 2theta "cos" (pi)/(3) = -"sin" 2theta`
`therefore Delta=|{:("sin"theta, "cos"theta, "sin" 2theta),(-"sin" theta,-"cos" theta, -"sin"2theta),("sin" (theta - (2pi)/(3)), "cos" (theta - (2pi)/(3)), "sin" (2theta -(4pi)/(3))):}|=0 " "["since," R_(1) "and "R_(2) "are proportional"]`