Question

If cos A+ cos ^2 A+ cos^3A=1 then find the value of sin^6A- 4sin^4A+ 8sin^2A

Answer 1

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

Consider,

$\red{\rm :\longmapsto\:cosA + {cos}^{2}A + {cos}^{3}A = 1}$

can be rewritten as

${\rm :\longmapsto\:cosA + {cos}^{3}A = 1 - {cos}^{2} A}$

We know,

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

So, using this, we get

${\rm :\longmapsto\:cosA + {cos}^{3}A = {sin}^{2} A}$

can be rewritten as

$\rm :\longmapsto\:cosA(1 + {cos}^{2}A) = {sin}^{2}A$

can be rewritten as

$\rm :\longmapsto\:cosA(1 + 1 - {sin}^{2}A) = {sin}^{2}A$

$\rm :\longmapsto\:cosA(2 - {sin}^{2}A) = {sin}^{2}A$

On squaring both sides, we get

$\rm :\longmapsto\: {cos}^{2}A {(2 - {sin}^{2} A)}^{2} = {sin}^{4}A$

can be further rewritten as using identity

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\rm :\longmapsto\: (1 - {sin}^{2}A) {(4 + {sin}^{4} A - 4 {sin}^{2}A )} = {sin}^{4}A$

$\rm :\longmapsto\: 4 + \cancel{ {sin}^{4} A} - 4 {sin}^{2}A - 4 {sin}^{2}A - {sin}^{6}A + {4sin}^{4}A = \cancel{ {sin}^{4}A}$

$\rm :\longmapsto\: 4 - 8 {sin}^{2}A - {sin}^{6}A + {4sin}^{4}A = 0$

$\rm :\longmapsto\: - 8 {sin}^{2}A - {sin}^{6}A + {4sin}^{4}A = - 4$

$\rm :\longmapsto\: 8 {sin}^{2}A + {sin}^{6}A - {4sin}^{4}A = 4$

$\bf\implies \: {sin}^{6}A - {4sin}^{4}A + {8sin}^{2}A = 4$

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