Question

sin^4a= 1/8(3-4cos2a+cos4a)​

Answer 1

Given :

$\\ \red \bigstar \large \rm \: {sin}^{4} a = \frac{1}{8(3 - 4cos2a + cos4a)}$

To prove :

LHS=RHS:

$\\ \red \bigstar \large \rm \: {sin}^{4}a = \frac{1}{8(3 - 4cos2a + cos4a)}$

Solution :

➡ LHS:

$\\ \implies \large \sf \: {sin}^{4} a = \bigg({ { \frac{2}{2} sin}^{2}a } \bigg)^{2} \\ \\ \\ \implies \large \sf \: \frac{1}{4} \bigg( {(1 - cos2x)}^{2} \bigg)\\ \\ \\ \implies \large \sf \: \frac{1}{4} \bigg(1 - 2cos2x + {cos}^{2} 2x \bigg) \\ \\ \\ \implies \large \sf \: \frac{2}{4 \times 2} \bigg(1 - 2cos2x + {cos}^{2} 2x \bigg) \\ \\ \\ \implies \large \sf \: \frac{ \cancel2}{4 \times \cancel2} \bigg(1 - 2cos2x + {cos}^{2} 2x \bigg) \\ \\ \\ \implies \large \sf \: \frac{1}{4} \bigg(1 - 2cos2x + {cos}^{2} 2x \bigg) \\ \\ \\ \implies \large \sf \: \frac{1}{8} \bigg(2 - 4cos2x + 2 {cos}^{2} 2x \bigg) \\ \\ \\ \implies \large \sf \: \frac{1}{8} \bigg(2 - 4cos2x + 1 + cos4x \bigg) \\ \\ \\ \implies \large \sf \: \frac{1}{8} \bigg(3 - 4cos2x + cos4x \bigg) \\ \\ \\ \implies \boxed{ \large \sf{ \therefore \: \frac{1}{8} \bigg(3 - 4cos2x + cos4x \bigg)}}$

LHS=RHS

hence proved..

Additional information :

$\large\bf \: (1) \: {sin}^{2} x + {cos}^{2} x = 1 \\ \\ \large\bf(2)1 + {tan}^{2} x = {sec}^{2} x \\ \\ \large \bf \: (3)1 + {cot}^{2} x = {cosec}^{2} x \\ \\ \large\bf (4) \: 1 - cos2x = 2 {sin}^{2}x \\ \\ \large\bf \: (5)1 + cos2x = 2 {sin}^{2} x \\ \\ \large\bf \: (6)1 - sin2x = ( {cosx - sinx})^{2} \\ \\ \large \bf \: (7)1 + sin2x = ({cosx + sinx})^{2} \\ \\ \large \bf \: (8) \: cos2x = {cos}^{2} x - {sin}^{2} x$