Provided, sin(A – B) = sinA cosB – cosA sinB, then sin15° will be
A.$\sqrt3-3\over2\sqrt3$
B.$\sqrt3-3\over\sqrt2$
C.$\sqrt3\over2\sqrt2$
D.$\sqrt3+3\over2\sqrt3$



[ A ]    a
right
[ B ]    b
[ C ]    c
[ D ]    d
Answer : Option A
Explanation :
sin (15º) = sin (45º – 30º)
= sin 45º cos30º – cos30º sin 45º
= $1\over\sqrt2$ x $\sqrt3\over2$ - $1\over2$x $1\over\sqrt2$
= $\sqrt3-1\over2\sqrt2$

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