Advertisements
Advertisements
प्रश्न
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
Advertisements
उत्तर
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
संबंधित प्रश्न
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Solve.
`sec75/(cosec15)`
Evaluate:
14 sin 30° + 6 cos 60° – 5 tan 45°
Use tables to find the acute angle θ, if the value of tan θ is 0.2419
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
Write the maximum and minimum values of sin θ.
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
`(sin 75^circ)/(cos 15^circ)` = ?
