Question

cos^6A+sin^6A=1-3sin^2Acos^2A​

Answer 1

GIVEN:

• cos⁶ A + sin⁶ A = 1 - 3sin² A cos² A

TO PROVE:

• cos⁶ A + sin⁶ A = 1 - 3sin² A cos² A

PROOF:

Take cos⁶ A + sin⁶ A as L.H.S

Take 1 - 3sin² A cos² A as R.H.S

${ \cos}^{6} A + { \sin}^{6} A = {( { \cos}^{2} A)}^{3} + {( { \sin}^{2} A)}^{3}$

A³ + B³ = (A + B)(A² - AB + B²)

${ \cos}^{2}A + { \sin}^{2} A({ \cos}^{4}A + { \sin}^{4} A - { \cos}^{2}A \: { \sin}^{2} A)$

Cos² A + Sin² A = 1

$1({ \cos}^{4}A + { \sin}^{4} A - { \cos}^{2}A \: { \sin}^{2} A)$

(A⁴ + B⁴ = (A² + B²)² - 2A²B²

cos⁴ A + sin⁴ A = (sin²A + cos²A)² - 2sin²Acos²A

cos⁴ A + sin⁴ A = 1² - 2sin²Acos²A

Sub. cos⁴A + sin⁴A = 1²- 2sin²Acos²A

$1 - 2{ \cos}^{2}A \: { \sin}^{2} A- { \cos}^{2}A \: { \sin}^{2} A$

$1 - 3{ \cos}^{2}A \: { \sin}^{2} A$

L.H.S = R.H.S

cos⁶ A + sin⁶ A = 1 - 3sin² A cos² A

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

sin^6A + cos^6A

= (sin^2A)3 + (cos^2A)3

= (sin^2A + cos^2A)(sin^4A – sin^2Acos^2A + cos^4A)

= 1 (sin^4A – sin^2Acos^2A + cos^4A + 2sin^2Acos^2A – 2sin^2Acos^2A)

= (sin2A + cos2A) – 3sin2Acos2A

= 1 – 3sin^2Acos^2A

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