Advertisements
Advertisements
प्रश्न
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
If x cos A + y sin A = m and x sin A – y cos A = n, then prove that : x2 + y2 = m2 + n2
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Define an identity.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
