Advertisements
Advertisements
Question
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
Advertisements
Solution
sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ)(sin4θ + cos2θ - sin2θcos2θ)
= 1 x [(sin2θ)2 + (cos2θ)2 + 2sin2θ.cos2θ - 3sin2θ.cos2θ]
= (sin2θ)2 + (cos2θ)2 - 3sin2θ.cos2θ
= 1 - 3sin2θ cos2θ.
= RHS
RELATED QUESTIONS
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
If (tan θ + sin θ) = m and (tan θ – sin θ) = n, prove that (m2 – n2)2 = 16 mn.
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
sin2θ + sin2(90 – θ) = ?
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
