2(sin⁶a+cos⁶a)-3(sin⁴a+cos⁴a)
Answers
✠ 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