Question

prove that:. sin⁶theta +cos⁶ theta = 1-3 sin² theta . cos²theta​

Answer 1

Step-by-step explanation:

LHS =(sin^2 theta + cos^2 theta)^3

=1

RHS

= sin^6 theta +cos^6 theta+3sin^2 theta ×cos^2 theta ( sin ^2 theta + cos ^2 theta)

= sin^6 theta +cos^6 theta+3sin^2 theta ×cos^2 theta

Taking 3sin^2 theta ×cos^2 theta from LHS to RHS

LHS = 1 -3sin^2 theta ×cos^2 theta

Now we have,

sin^6 theta +cos^6 theta = 1 -3sin^2 theta ×cos^2 theta

( as we have just transposed the terms using ( a+b)^3 = a^3+ b ^3 + 3 ab ( a + b ) identity....

Hence proved

Answer 2

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

$\sf \: Prove \: that : \: {sin}^{6} \theta \: + \: {cos}^{6} \theta = 1 - 3 {sin}^{2} \theta \: {cos}^{2} \theta$

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

We know,

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

and

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

Now,

Consider,

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

can be rewritten as

$\sf \: \: \: \: = \: \: {\bigg( {sin}^{2} \theta \bigg) }^{3} + { \bigg( {cos}^{2} \theta\bigg) }^{3}$

$\sf \: \: \: \: = \: \: {\bigg({sin}^{2} \theta + {cos}^{2} \theta \bigg) }^{3} - 3 {sin}^{2} \theta \: {cos}^{2} \theta \: \bigg( {sin}^{2} \theta + {cos}^{2} \theta\bigg)$

$\sf \: \: \: \: = \: \: {(1)}^{3} - 3 \: {sin}^{2} \theta \: {cos}^{2} \theta \times 1$

$\sf \: \: \: \: = \: \: 1 - 3 \: {sin}^{2} \theta \: {cos}^{2} \theta$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

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