Prove the following identity : sin^ 6 A+cos^ 6 A=1-3 sin^ 2 A cos^ 2 A
Answers
Step-by-step explanation:
Given :-
Sin⁶ A + Cos⁶ A
To find:-
Prove that :
Sin⁶ A + Cos⁶ A = 1-3 Sin² A Cos² A.
Solution :-
On taking LHS :
Sin⁶ A + Cos⁶ A
It can be written as
=> (Sin² A)³ + (Cos² A)³
It is in the form of a³+b³
Where , a = Sin² A and b = Cos² A
We know that
a³ + b³ = (a+b)(a²-ab+b²)
=>(Sin² A+Cos² A)[(Sin² A)²-Sin²ACos²A+ (Cos² A)²]
(Sin²A+Cos²A)(Sin⁴A-Sin²ACos²A+Cos⁴ A)
We know that
Sin² A + Cos² A = 1
=> (1)(Sin⁴ A-Sin²ACos²A+ Cos⁴ A)
=>Sin⁴ A-Sin²ACos²A+ Cos⁴ A
=> Sin⁴ A +Cos⁴ A -Sin²A Cos²A
=> (Sin² A)²+(Cos² A)² -Sin²A Cos²A
We know that
a²+b² = (a+b)²-2ab
So , Sin⁴A+Cos⁴A = (Sin²A+Cos²A)²-2Sin²ACos²A
=> (Sin² A+ Cos² A)²-2 Sin² A Cos² A-Sin²A Cos²A
=>(Sin² A+ Cos² A)²-3 Sin² A Cos² A
We know that
Sin² A + Cos² A = 1
=> (1)² - 3 Sin² A Cos² A
=> 1 - 3 Sin² A Cos² A
=> RHS
=> LHS = RHS
Answer:-
Sin⁶ A + Cos⁶ A = 1 - 3 Sin² A Cos² A
Used formulae:-
- Sin² A + Cos² A = 1
- a³ + b³ = (a+b)(a²-ab+b²)