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Have a tough question .I expect an answer.If sinA+sin2A+sin3A = 1.then prove that .cos6A-4cos4A+8cos2A=4.the replier has his trigonometry almost perfect
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Have a tough question .I expect an answer.If sinA+sin2A+sin3A = 1.then...
Let me ,make it simple ,bro has given good answer ,but square looks like "2" so I will re-write .sinA + sin²A + sin³A = 1=> sinA + sin³A = 1 - sin²A=> sinA ( 1+sin²A ) = cos​²A=> sinA ( 1 + 1 - cos²A ) = cos²A=> sinA ( 2 - cos²A ) = cos²A=> sinA = cos²A / ( 2 - cos²A)=> √ ( 1 - cos²A ) = cos²A / 2- cos²A=> 1 - cos²A = ( cos²A )² /  ( 2 - cos²A )²=> 1 - cos²A = cos⁴A / ( 4 + cos⁴A - 4 cos²A )=>  ( 1 - cos²A ) ( 4 + cos⁴A - 4 cos²A ) = cos⁴A=>  ( ​4 + cos²A - 4 cos²A -4cos²A - cos^6A + 4cos⁴A = cos⁴A=>  ( 4 -8cos²A - cos^6A + 4cos⁴A ) = cos⁴A - cos⁴A=>  ( 4 -8cos²A - cos^6A + 4 cos⁴A ) = 0=>  4 = 8cos²A + cos^6A - 4cos⁴A
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Have a tough question .I expect an answer.If sinA+sin2A+sin3A = 1.then...
SinA + sin2A + sin3A = 1=> sinA + sin3A = 1 - sin2A=> sinA ( 1+sin2A ) = cos​2A=> sinA ( 1 + 1 - cos2A ) = cos2A=> sinA ( 2 - cos2A ) = cos2A=> sinA = cos2A / ( 2 - cos2A)=> sqrt ( 1 - cos2A ) = cos2A / 2- cos2A=> 1 - cos2A = ( cos2A )2 /  ( 2 - cos2A )2=> 1 - cos2A = cos4A / ( 4 + cos4A - 4 cos2A )=>  ( 1 - cos2A ) ( 4 + cos4A - 4 cos2A ) = cos4A=>  ( ​4 + cos4A - 4 cos2A -4cos​2A - cos6A + 4cos4A = cos4A=>  ( 4 -8cos2A - cos6A + 4cos4A ) = cos4A - cos4A=>  ( 4 -8cos2A - cos6A + 4 cos4A ) = 0=>  4 = 8cos2A + cos6A - 4cos4A
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Have a tough question .I expect an answer.If sinA+sin2A+sin3A = 1.then prove that .cos6A-4cos4A+8cos2A=4.the replier has his trigonometry almost perfect
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