if sin theta + sin square theta equals to 1 find the value of cos 12 theta + 3 cos tan theta + 3 cos 8 theta + cos 6 theta + 2 cos 4 theta + 2 cos 2 theta minus 2
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sinx +sin²x=1
⇒sin²x=1-sinx
⇒1-cos²x=1-sinx
⇒cos²x=sinx
now put the value of cos²x you will get the answer easily.
[tex]cos^{6} x(cos ^{6} x+3cos ^{4} x+3cos ^{2} x+1)-1 =cos ^{6} x(cos ^{2} x+1) ^{3} -1 [/tex]
=(sin^2x+sinx)^3-1=1-1=0
⇒sin²x=1-sinx
⇒1-cos²x=1-sinx
⇒cos²x=sinx
now put the value of cos²x you will get the answer easily.
[tex]cos^{6} x(cos ^{6} x+3cos ^{4} x+3cos ^{2} x+1)-1 =cos ^{6} x(cos ^{2} x+1) ^{3} -1 [/tex]
=(sin^2x+sinx)^3-1=1-1=0
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