Advertisements
Advertisements
Question
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
Advertisements
Solution
L.H.S. = cos 3A = cos 90° = 0
R.H.S. = 4 cos3 A – 3 cos A
= 4 cos3 30° – 3 cos 30°
= `4 (sqrt3/2)^3 - 3(sqrt3/2)`
= `(4 xx 3sqrt(3))/8 - (3sqrt(3))/2`
= `(3sqrt(3))/2 - (3sqrt(3))/2 = 0`
L.H.S. = R.H.S.
APPEARS IN
RELATED QUESTIONS
If sin θ =3/5, where θ is an acute angle, find the value of cos θ.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate:
cosec (65° + A) – sec (25° – A)
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
If ∠A = 30°, then tan 2A = ?
`(sin 75^circ)/(cos 15^circ)` = ?
