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 the angle θ= –60º, find the value of cosθ.
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Evaluate cosec 31° − sec 59°
if `tan theta = 3/4`, find the value of `(1 - cos theta)/(1 +cos theta)`
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Use tables to find sine of 21°
If tanθ = 2, find the values of other trigonometric ratios.
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
