Advertisements
Advertisements
Question
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
Advertisements
Solution
4 cos2 A − 3 = 0
`cos A = sqrt(3)/2`
We know `cos 30^circ = sqrt(3)/2`
So, A = 30°
L.H.S. = sin 3A = sin 90° = 1
R.H.S. = 3 sin A – 4 sin3 A
= 3 sin 30° – 4 sin3 30°
= `3 xx 1/2 - 4 xx (1/2)^3` ...{∵ sin 30° = `1/2`}
= `3/2 - 4 xx 1/8`
= `3 /2 - 1/2`
= `2/2`
= 1
L.H.S. = R.H.S.
APPEARS IN
RELATED QUESTIONS
Without using trigonometric tables evaluate:
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
Write the value of tan 10° tan 15° tan 75° tan 80°?
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to ______.
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?
If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to ______.
