Question

2(sin⁶a+cos⁶a)-3(sin⁴a+cos⁴a)

Answer 1

✠ Given to find the value of :-

2 (sin⁶A + cos⁶A) -3 (sin⁴A + cos⁴A)

✠Answer :-

- 1

✠SOLUTION:-

⇢sin⁶A + cos⁶A can be written as (sin²A)³ + (cos²A)³

⇢ sin⁴A + cos⁴A can be written as (sin²A)² + (cos²A)²

= 2 (sin⁶A + cos⁶A) -3 (sin⁴A + cos⁴A)

= 2[ (sin²A)³ + (cos²A)³] -3[ (sin²A)² + (cos²A)²]

✠As we know that ,

a³ + b³ = (a+b)³ -3ab(a+b)

a² + b² = (a+b)² 2ab

✠ By applying these formulae we get ,

2[ (sin²A)³ + (cos²A)³] -3[ (sin²A)² + (cos²A)²]

= 2[(sin²A+cos²A)³ -3 sin²Acos²A(sin²A+cos²A)] -3 [(sin²A+cos²A)² -2 sin²Acos²A]

✠From Trigonmetric Identities,

sin²A + cos²A = 1

= 2[(1)³ -3 sin²Acos²A(1)] -3 [(1)² -2 sin²Acos²A]

= 2[ 1- 3sin²Acos²A] -3 [ 1- 2sin²Acos²A]

= 2- 6 sin²Acos²A - 3 + 6 sin²Acos²A

= 2-3 -6sin²Acos²A + 6sin²Acos²A

= 2-3

= - 1

✠So, the value of 2 (sin⁶A + cos⁶A) -3 (sin⁴A + cos⁴A)

is - 1

✠Used formulae:-

✠ Algebraic identities :-

⇢a³ + b³ = (a+b)³ -3ab(a+b)

⇢a² + b² = (a+b)² 2ab

✠Trigonmetric Identities:-

⇢sin²A + cos²A = 1

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✠Know more some Algebraic identities:-

⇢(a+ b)² = a² + b² + 2ab

⇢( a - b )² = a² + b² - 2ab

⇢( a + b )² + ( a - b)² = 2a² + 2b²

⇢( a + b )² - ( a - b)² = 4ab

⇢( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

⇢a² + b² = ( a + b)² - 2ab

⇢(a + b )³ = a³ + b³ + 3ab ( a + b)

⇢( a - b)³ = a³ - b³ - 3ab ( a - b)

⇢If a + b + c = 0 then a³ + b³ + c³ = 3abc

✠Know more some Trigonmetric relations, ratios , identities:-

✠Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

✠Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

✠Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj