Question

Prove the following identity : sin^ 6 A+cos^ 6 A=1-3 sin^ 2 A cos^ 2 A​

Answer 1

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)