Question

(Cosa+sina)^4-(cosa-sina)^4=8cosa sina

Answer 1

$\bold{ \underline{\red{To \: prove}}} \longrightarrow \\ {(cos \alpha + sin \alpha )}^{4} - {(cos \alpha - sin \alpha )}^{4} = 8cos \alpha \: sin \alpha \\ \bold{ \underline{\red{Concept \: used}}} \longrightarrow \\ \boxed{\pink {(x + y)^{2} = {x}^{2} + {y}^{2} + 2xy}} \\ \boxed{ \pink{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}} \\ \boxed{ \pink{ {x}^{2} - {y}^{2} = (x + y) \:( x - y)}} \\ \bold { \underline{\red{Solution}}} \longrightarrow \\ \blue{LHS} \\ = {(cos \alpha + sin \alpha) }^{4} - {(cos \alpha - sin \alpha }^{4} \\ = { ({(cos \alpha + sin \alpha) }^{2}) }^{2} - {( {(cos \alpha - sin \alpha) }^{2} )}^{2} \\ = {( {cos}^{2} \alpha + {sin}^{2} \alpha + 2cos \alpha sin \alpha )}^{2} - {( {cos}^{2} \alpha + {sin}^{2} \alpha - 2sin \alpha cos \alpha ) }^{2} \\ = {(1 + 2sin \alpha cos \alpha) }^{2} - (1 - 2sin \alpha cos \alpha ) \\ now \: applying \: {x}^{2} - {y}^{2} = (x + y)(x - y) \\ = (1 + 2sin \alpha cos \alpha + 1 - 2sin \alpha cos \alpha ) \: (1 + 2sin \alpha cos \alpha - (1 - 2sin \alpha cos \alpha ) \\ = (2) \: (1 + 2sin \alpha cos \alpha - 1 + 2sin \alpha cos \alpha ) \\ = 2 \: (4sin \alpha cos \alpha ) \\ = 8 \: sin \alpha \: cos \alpha \\ =\blue{ RHS}$

Answer 2

Step-by-step explanation:

$\begin{gathered} \bold{ \underline{\red{To \: prove}}} \longrightarrow \\ {(cos \alpha + sin \alpha )}^{4} - {(cos \alpha - sin \alpha )}^{4} = 8cos \alpha \: sin \alpha \\ \bold{ \underline{\red{Concept \: used}}} \longrightarrow \\ \boxed{\pink {(x + y)^{2} = {x}^{2} + {y}^{2} + 2xy}} \\ \boxed{ \pink{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}} \\ \boxed{ \pink{ {x}^{2} - {y}^{2} = (x + y) \:( x - y)}} \\ \bold { \underline{\red{Solution}}} \longrightarrow \\ \blue{LHS} \\ = {(cos \alpha + sin \alpha) }^{4} - {(cos \alpha - sin \alpha }^{4} \\ = { ({(cos \alpha + sin \alpha) }^{2}) }^{2} - {( {(cos \alpha - sin \alpha) }^{2} )}^{2} \\ = {( {cos}^{2} \alpha + {sin}^{2} \alpha + 2cos \alpha sin \alpha )}^{2} - {( {cos}^{2} \alpha + {sin}^{2} \alpha - 2sin \alpha cos \alpha ) }^{2} \\ = {(1 + 2sin \alpha cos \alpha) }^{2} - (1 - 2sin \alpha cos \alpha ) \\ now \: applying \: {x}^{2} - {y}^{2} = (x + y)(x - y) \\ = (1 + 2sin \alpha cos \alpha + 1 - 2sin \alpha cos \alpha ) \: (1 + 2sin \alpha cos \alpha - (1 - 2sin \alpha cos \alpha ) \\ = (2) \: (1 + 2sin \alpha cos \alpha - 1 + 2sin \alpha cos \alpha ) \\ = 2 \: (4sin \alpha cos \alpha ) \\ = 8 \: sin \alpha \: cos \alpha \\ =\blue{ RHS}\end{gathered}$

✓HENCE PROVED ✔️