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.
if `tan theta = 12/5` find the value of `(1 + sin theta)/(1 -sin theta)`
Solve.
`tan47/cot43`
Use tables to find cosine of 8° 12’
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
Evaluate:
`2(tan35^@/cot55^@)^2 + (cot55^@/tan35^@)^2 - 3(sec40^@/(cosec50^@))`
The value of
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
