[tex]\large\underline{\sf{Solution-}}[/tex] Given that [tex]\rm :\longmapsto\:cos\phi + {cos}^{2} \phi = 1[/tex] can be rewritten as [tex]\rm :\longmapsto\:cos\phi = 1 – {cos}^{2} \phi [/tex] [tex]\rm :\longmapsto\:cos\phi = {sin}^{2} \phi [/tex] [tex]\red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}[/tex] Consider, [tex]\rm :\longmapsto\: {sin}^{12}\phi + 3 {sin}^{10}\phi + 3 {sin}^{8}\phi + {sin}^{6}\phi [/tex] [tex]\rm \: = \: \: {\bigg( {sin}^{4}\phi \bigg) }^{3} + 3 {sin}^{6}\phi \bigg( {sin}^{4}\phi + {sin}^{2}\phi\bigg) + {\bigg( {sin}^{2}\phi \bigg) }^{3}[/tex] We know, [tex] \boxed{ \sf \: {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}[/tex] So, using this identity, we get [tex]\rm \: = \: \:{\bigg( {sin}^{4}\phi + {sin}^{2}\phi \bigg) }^{3}[/tex] [tex]\rm \: = \: \:{\bigg( ({sin}^{2}\phi)^{2} + {sin}^{2}\phi \bigg) }^{3}[/tex] [tex]\rm \: = \: \:{\bigg( {cos}^{2}\phi + cos\phi \bigg) }^{3}[/tex] [tex] \blue{\bf:\longmapsto\: \: \: \: \: \: \: \: \because \: cos\phi = {sin}^{2} \phi }[/tex] [tex]\rm \: = \: \: {(1)}^{3} [/tex] [tex] \blue{\bf:\longmapsto\: \: \: \: \: \: \: \: \because \: cos\phi + {cos}^{2} \phi = 1}[/tex] Hence, [tex] \purple{ \boxed{\bf :\longmapsto\: {sin}^{12}\phi + 3 {sin}^{10}\phi + 3 {sin}^{8}\phi + {sin}^{6}\phi = 1}}[/tex] 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 Reply
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\rm :\longmapsto\:cos\phi + {cos}^{2} \phi = 1[/tex]
can be rewritten as
[tex]\rm :\longmapsto\:cos\phi = 1 – {cos}^{2} \phi [/tex]
[tex]\rm :\longmapsto\:cos\phi = {sin}^{2} \phi [/tex]
[tex]\red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}[/tex]
Consider,
[tex]\rm :\longmapsto\: {sin}^{12}\phi + 3 {sin}^{10}\phi + 3 {sin}^{8}\phi + {sin}^{6}\phi [/tex]
[tex]\rm \: = \: \: {\bigg( {sin}^{4}\phi \bigg) }^{3} + 3 {sin}^{6}\phi \bigg( {sin}^{4}\phi + {sin}^{2}\phi\bigg) + {\bigg( {sin}^{2}\phi \bigg) }^{3}[/tex]
We know,
[tex] \boxed{ \sf \: {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}[/tex]
So, using this identity, we get
[tex]\rm \: = \: \:{\bigg( {sin}^{4}\phi + {sin}^{2}\phi \bigg) }^{3}[/tex]
[tex]\rm \: = \: \:{\bigg( ({sin}^{2}\phi)^{2} + {sin}^{2}\phi \bigg) }^{3}[/tex]
[tex]\rm \: = \: \:{\bigg( {cos}^{2}\phi + cos\phi \bigg) }^{3}[/tex]
[tex] \blue{\bf:\longmapsto\: \: \: \: \: \: \: \: \because \: cos\phi = {sin}^{2} \phi }[/tex]
[tex]\rm \: = \: \: {(1)}^{3} [/tex]
[tex] \blue{\bf:\longmapsto\: \: \: \: \: \: \: \: \because \: cos\phi + {cos}^{2} \phi = 1}[/tex]
Hence,
[tex] \purple{ \boxed{\bf :\longmapsto\: {sin}^{12}\phi + 3 {sin}^{10}\phi + 3 {sin}^{8}\phi + {sin}^{6}\phi = 1}}[/tex]
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