Advertisements
Advertisements
Question
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
Solution
LHS = `sin^8θ - cos^8θ`
= `(sin^4θ)^2 - (cos^4θ)^2`
= `(sin^4θ - cos^4θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)(sin^2θ + cos^2θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)(sin^4θ + cos^4θ)`
= `(sin^2θ - cos^2θ)((sin^2θ)^2 + (cos^2θ)^2 + 2sin^2θcos^2θ - 2sin^2θcos^2θ)`
= `(sin^2θ - cos^2θ)((sin^2θ + cos^2θ)^2 - 2sin^2θcos^2θ)`
= `(sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
If tan θ × A = sin θ, then A = ?
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.