Question - Prove that: $cos^6A + sin^6A =$
(i) $(1 - sin^2A cos^2A)$
(ii) $\left( 1 - \dfrac{sin^2 2A}{4} \right)$
(iii) $\dfrac{4 - sin^22A}{4}$

This is a class 10 Question From Multiple Angles chapter of Unit Trigonometry.

Solution:

Taking LHS
$= cos^6 A + sin^6A$
$= (cos^2 A)^3 + (sin^2 A)^2$
$= (cos^2 A + sin^2 A) (cos^4A + cos^2A sin^2A + sin^4A)$

$= 1 ( cos^4 A + sin^4A + cos^2A sin^2A)$

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

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

$= (1 - sin^2A cos^2A)$ - (i)
$= \left(1 - \dfrac{4 sin^2A cos^2A}{4} \right)$

$= \left( 1 - \dfrac{sin^2 2A}{4} \right)$ -(ii)

$= \dfrac{4 - sin^22A}{4}$ -(iii)

#proved

Identities Used:
• $cos^2A + sin^2A = 1$
• $2sinAcosA = sin2A$
Formulae Used:
• (a³ + b³) = (a+b)(a² - ab + b²)

Related Solution:

Related Notes:
Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

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