Advertisements
Advertisements
Question
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
Advertisements
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.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
sec4 A − sec2 A is equal to
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
