how to show that csc⁡x−csc⁡(π3+x)+csc⁡(π3−x)=3csc⁡3x? My attempt: LHS=csc⁡x−csc⁡(π3+x)+csc⁡(π3−x) =1sin⁡x−1sin⁡(π3+x)+1sin⁡(π3−x) =sin⁡xsin⁡(π3+x)+sin⁡(π3+x)sin⁡(π3−x)−sin⁡(π3−x)sin⁡xsin⁡xsin⁡(π3+x)sin⁡(π3−x) =4sin⁡3x(sin⁡xsin⁡(π3+x)+sin⁡(π3+x)sin⁡(π3−x)−sin⁡(π3)sin⁡x) =4sin⁡3x(sin⁡xsin⁡(π3+x)−sin⁡(π3−x)sin⁡(sin⁡x−sin⁡(π3+x))) =4sin⁡3x(sin⁡xsin⁡(π3+x)−sin⁡(π3−x)(2sin⁡−π6cos⁡(x+π6)))

Question

how to show that csc⁡x−csc⁡(π3+x)+csc⁡(π3−x)=3csc⁡3x?  My attempt:   LHS=csc⁡x−csc⁡(π3+x)+csc⁡(π3−x)   =1sin⁡x−1sin⁡(π3+x)+1sin⁡(π3−x)   =sin⁡xsin⁡(π3+x)+sin⁡(π3+x)sin⁡(π3−x)−sin⁡(π3−x)sin⁡xsin⁡xsin⁡(π3+x)sin⁡(π3−x)  =4sin⁡3x(sin⁡xsin⁡(π3+x)+sin⁡(π3+x)sin⁡(π3−x)−sin⁡(π3)sin⁡x)  =4sin⁡3x(sin⁡xsin⁡(π3+x)−sin⁡(π3−x)sin⁡(sin⁡x−sin⁡(π3+x)))   =4sin⁡3x(sin⁡xsin⁡(π3+x)−sin⁡(π3−x)(2sin⁡−π6cos⁡(x+π6)))

Jordan Mitchell · Accepted Answer

We have  sin⁡xsin⁡(π3+x)+sin⁡(π3−x)cos⁡(x+π6)   =sin⁡(x)[sin⁡(π3)cos⁡(x)+cos⁡(π3)sin⁡(x)]+(sin⁡(π3)cos⁡(x)−cos⁡(π3)sin⁡(x))(cos⁡(π6)cos⁡x−sin⁡(π6)sin⁡(x))  =sin⁡(x)[32cos⁡(x)+12sin⁡(x)]+(32cos⁡(x)−12sin⁡(x))(32cos⁡(x)−12sin⁡(x))  =32sin⁡(x)cos⁡(x)+12sin2x+34cos2(x)−32cos⁡(x)sin⁡(x)+14sin2(x)  or  34(sin2(x)+cos2(x))=34

Mason Hall · Accepted Answer

1sin⁡x−1sin⁡(π3+x)+1sin⁡({{−x)=  =sin⁡(π3−x)sin⁡(π3+x)+sin⁡x(sin⁡(π3+x)−sin⁡(π3−x))sin⁡xsin⁡(π3+x)sin⁡(π3−x)  =12(cos⁡2x−cos⁡2π3)+sin⁡x⋅2sin⁡xcos⁡π3sin⁡x(32cos⁡x+12sin⁡x)(32cos⁡x−12sin⁡x)  =12−sin2x+14+sin2xsin⁡x(34cos2x−14sin2x)=3sin⁡x(3(1−sin2x)−sin2x)=3sin⁡3x

Vasquez · Accepted Answer

Using the identitiessin⁡Asin⁡B=12(cos⁡(A−B)−cos⁡(A+B))sin⁡Acos⁡B=12(sin⁡(A+B)+sin⁡(A−B))we havesin⁡xsin⁡(π3+x)+sin⁡(π3−x)cos⁡(x+π6)=12(cos⁡(−π3)−cos⁡(2x+π3)+sin⁡π2+sin⁡(π6−2x))=12(12+1−cos⁡(2x+π3)+cos⁡(2x+π3))=34

how to show that \csc x-\csc(\frac{\pi}{3}+x)+\csc(\frac{\pi}{3}-x)=3 \csc 3x? My attempt: LHS=\csc x−\csc(\frac{\pi}{3}+x)+\csc(\frac{\pi}{3}-x) =\frac{1}{\sin

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