Question

Prove:
$\cos(x) ^{6} + \sin(x)^{6} = 1 - 3 \cos(x)^{2} - 3 \cos(x)^{4}$
Who spams is a fool.​

Answer 1

Correct question is :

$\sf \: prove : \\ {\rm \: {cos}^{6} x + {sin}^{6} x = 1 - 3 {cos}^{2} x + 3 {cos}^{4} x}$

Solution :

$\orange{ \boxed{\boxed{\begin{array}{cc}\rm \: L.H.S = {cos}^{6}x + {sin}^{6} x \\ \\ \rm = {( {cos}^{2}x) }^{3} + {( {sin}^{2} x)}^{3} \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \to \:we \: know \: that \\ \\ \rm \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \end{array}}}} \\ \\ \small{ \rm = {( {cos}^{2}x + {sin}^{2} x)}^{3} - 3 {cos}^{2}x. {sin}^{2} x( {cos}^{2} x + {sin}^{2}x)} </p><p>\\ \\ \rm = {1}^{3} - 3 {cos}^{2} x. {sin}^{2} x \times 1 \\ \\ \rm = 1 - 3 {cos}^{2}x(1 - {cos}^{2} x) \\ \\ \rm = 1 - 3 {cos}^{2} x + 3 {cos}^{4}x \\ \\ = \rm \: R.H.S \end{array}}}}$

Answer 2

Appropriate Question :-

Prove that

$\rm :\longmapsto\: {sin}^{6}x \: + \: {cos}^{6}x = 1 - 3 {cos}^{2}x + 3{cos}^{4}x$

$\purple{\large\underline{\sf{Solution-}}}$

Consider LHS

$\rm :\longmapsto\: {sin}^{6}x \: + \: {cos}^{6}x$

can be rewritten as

$\rm \: = \: {\bigg[ {sin}^{2}x\bigg]}^{3} + {\bigg[ {cos}^{2}x \bigg]}^{3}$

We know,

$\red{\boxed{ \rm{ \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \: }}}$

So, using this identity, we get

$\rm \: = \: {\bigg[ {sin}^{2}x + {cos}^{2}x \bigg]}^{3} - 3 {sin}^{2}x {cos}^{2}x\bigg[ {sin}^{2}x + {cos}^{2}x \bigg]$

We know,

$\red{\boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1 \: \: }}}$

So, using this identity, we get

$\rm \: = \: {1}^{3} - 3 {sin}^{2}x {cos}^{2}x(1)$

$\rm \: = \: 1 - 3 {sin}^{2}x {cos}^{2}x$

can be rewritten as

$\rm \: = \: 1 - 3 \: [1 - {cos}^{2}x ] \: {cos}^{2}x$

$\rm \: = \:1 - 3 {cos}^{2}x + 3{cos}^{4}x$

Hence,

$\red{\rm :\longmapsto\:} \red{\boxed{ \rm{ \: {sin}^{6}x \: + \: {cos}^{6}x = 1 - 3 {cos}^{2}x + 3{cos}^{4}x \: \: }}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1