Question

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sin^6+cos^6=1-3sin^2cos^2​

Answer 1

SOLUTION

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Answer 2

Sin6A+Cos6A  

[Sin ^{2} A]^3 + [Cos ^{2}]^3[Sin2A]3+[Cos2]3                [.·. (a³+b³)= (a+b) (a²-ab+b²) ]

⇒(Cos^{2} A + Sin^{2} A) (Cos^{4} A-Cos^{2} A. Sin^{2} A+ Sin^{4} A)(Cos2A+Sin2A)(Cos4A−Cos2A.Sin2A+Sin4A)

⇒(Sin^{4} A + Cos^{4} A) - Sin^{2} A.Cos^{2} A(Sin4A+Cos4A)−Sin2A.Cos2A 

⇒(Sin^{2}A+Cos^{2}A)^{2} -2Sin^{2}A.Cos ^{2}A-Sin^{2}A.Cos^{2}A(Sin2A+Cos2A)2−2Sin2A.Cos2A−Sin2A.Cos2A

[.·. Cos²A+Sin²A=1]

⇒1-3Sin^{2}A.Cos^{2}ASin2A.Cos2A              

[Hence, proved]

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