Question

prove that sin⁴A - cos⁴A = 2sin²A - 1​

Answer 1

${EXPLANATION}$

$\maltese$Given to prove :-

sin⁴A - cos⁴A = 2sin²A-1

$\maltese$Solution:-

Take L.H.S that is

sin⁴A - cos⁴A

sin⁴ A can be written as (sin²A)²

Similarily,

cos⁴ A can be written as (cos²A)²

So, substitute the values,

= (sin²A)² - (cos²A)²

It is in form of a² -b² = (a+b)(a-b)

So,

(sin²A +cos²A)(sin²A-cos²A)

From trigonometric identities We know that ,

sin²A + cos²A = 1 So,

1(sin²A -cos²A)

sin²A - cos²A ------ eq 1

So, we know that ,

sin²A + cos²A = 1

cos²A = 1-sin²A ------ substitute this value in eq 1

sin²A - (1-sin²A)

sin²A - 1 + sin²A

2sin²A - 1

Hence , proved !

$\maltese$USED FORMULAE:-

$\maltese$Trigonmetric identity :-

• sin²A + cos²A = 1
• sin²A = 1-cos²A

$\maltese$Algebraic Identitiy

• (a+b)(a-b) = a²-b²

$\maltese$Know more :-

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