Advertisements
Advertisements
Question
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Advertisements
Solution
sin θ (1 + sin2 θ) = cos2 θ
sin θ (1 + 1 – cos2 θ) = cos2 θ
sin θ (2 – cos2 θ) = cos2 θ
Squaring on both sides,
sin2 θ (2 – cos2 θ)2 = cos4 θ
(1 – cos2 θ) (4 + cos4 θ – 4 cos2 θ) = cos4 θ
4 cos4 θ – 4 cos2 θ – cos6 θ + 4 cos4 θ = cos4 θ
4 + 5 cos4 θ – 8 cos2 θ – cos6 θ = cos4 θ
– cos6 θ + 5cos4 θ – cos4 θ – 8 cos2 θ = – 4
– cos6 θ + 4cos4 θ – 8 cos2 θ = – 4
cos6 θ – 4cos4 θ + 8 cos2 θ = 4
Hence it is proved
APPEARS IN
RELATED QUESTIONS
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
If ` cot A= 4/3 and (A+ B) = 90° ` ,what is the value of tan B?
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
If 2sin2β − cos2β = 2, then β is ______.
