Question

Prove that Sin⁶theta by 2+ Cos⁶theta by 2 = 1 by4 ( 1 + 3 cos²theta).
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Answer 1

Step-by-step explanation:

${sin}^{6} \frac{ \theta}{2} + {cos}^{6} \frac{ \theta}{2} \\ = { ({sin}^{2} \frac{ \theta}{2} })^{3} + ( { {cos}^{2} \frac{ \theta}{2} })^{3} \\ = ( {sin}^{2} \frac{ \theta}{2} + {cos}^{2} \frac{ \theta}{2} )( {sin}^{4} \frac{ \theta}{2} + {cos}^{4} \frac{ \theta}{2} - sin ^{2} \frac{ \theta}{2} cos ^{2} \frac{ \theta}{2} ) \\ = {sin}^{4} \frac{ \theta}{2} + {cos}^{4} \frac{ \theta}{2} - {sin}^{2} \frac{ \theta}{2} {cos}^{2} \frac{ \theta}{2} \\ = ( { {cos}^{2} \frac{ \theta}{2} - {sin}^{2} \frac{ \theta}{2} })^{2} + sin ^{2} \frac{ \theta}{2} {cos}^{2} \frac{ \theta}{2} \\ = (cos2. \frac{ \theta}{2} ) ^{2} + \frac{1}{4} ( {2sin \frac{ \theta}{2}cos \frac{ \theta}{2} })^{2} \\ = {cos}^{2} \theta + \frac{1}{4} sin ^{2} \theta \\ = {cos}^{2} \theta \: + \frac{1}{4} - \frac{1}{4} {cos}^{2} \theta \\ = \frac{1}{4} + ( {cos}^{2} \theta - \frac{1}{4} {cos}^{2} \theta) \\ = \frac{1}{4} + \frac{3 {cos}^{2} \theta }{4} \\ = \frac{1}{4} (1 + 3 {cos}^{2} \theta)$