Advertisements
Advertisements
प्रश्न
If sin A = `1/3`, sin B = `1/4` then find the value of sin (A + B) where A and B are acute angles.
Advertisements
उत्तर
Since A and B are acute angles, all the ratios are positive.
sin A = `1/3`; sin B = `1/4`; cos A = `(2sqrt2)/3`; cos B = `sqrt15/4`
∴ sin (A + B) = sin A cos B + cos A sin B
`= 1/3 * sqrt15/4 + (2sqrt2)/3 * 1/4`
`= sqrt15/12 + (2sqrt2)/12`
`= (sqrt15 + 2sqrt2)/12`
APPEARS IN
संबंधित प्रश्न
Find the value of the following:
cosec 15º
Find the value of the following:
cos 70° cos 10° – sin 70° sin 10°
Prove that:
tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Prove that `(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
If tan x = `3/4` and `pi < x < (3pi)/2`, then find the value of sin `x/2` and cos `x/2`.
Prove that `2 sin^2 (3pi)/4 + 2 cos^2 pi/4 + 2 sec^2 pi/3` = 10
The value of cos(-480°) is:
The value of sin 28° cos 17° + cos 28° sin 17°
The value of 1 – 2 sin2 45° is:
The value of 4 cos3 40° – 3 cos 40° is
