If sina + sin^2a + sin^3a = 1 then prove that cos^6a + 4cos^4a + 8cos^

1.	If sina + sin^2a + sin^3a = 1 then prove that cos^6a + 4cos^4a + 8cos^2a = 4
Answer» We have,sin{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0+ sin3{tex}\\theta{/tex}\xa0= 1{tex}\\Rightarrow{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ sin3{tex}\\theta{/tex}\xa0= 1 - sin2{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0sin{tex}\\theta{/tex}\xa0(1 + sin2{tex}\\theta{/tex}) = cos2{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0sin2{tex}\\theta{/tex}\xa0(1 + sin2{tex}\\theta{/tex})2 = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}\xa0){\xa01 + (1 - cos2{tex}\\theta{/tex}\xa0) }2\xa0= cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}) (2 - cos2{tex}\\theta{/tex})2 = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - cos2{tex}\\theta{/tex}) (4 - 4 cos2{tex}\\theta{/tex}\xa0+ cos4{tex}\\theta{/tex}) = cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa04 - 4 cos2{tex}\\theta{/tex}\xa0+ cos4{tex}\\theta{/tex}\xa0- 4 cos2{tex}\\theta{/tex}\xa0+ 4 cos4{tex}\\theta{/tex}\xa0- cos6{tex}\\theta{/tex}\xa0= cos4{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0-cos6{tex}\\theta{/tex}\xa0+ 4 cos4{tex}\\theta{/tex}\xa0- 8cos2{tex}\\theta{/tex}\xa0+ 4 = 0\xa0{tex}\\Rightarrow{/tex}\xa0cos6{tex}\\theta{/tex}\xa0- 4cos4{tex}\\theta{/tex}\xa0+ 8cos2{tex}\\theta{/tex}\xa0= 4

If sina + sin^2a + sin^3a = 1 then prove that cos^6a + 4cos^4a + 8cos^2a = 4