if A,B and A+B are positive acute angles , prove that
sin(A+B) = sin A . cos B + cos A . sin B
Answers
Step-by-step explanation:
ANSWER
3
2sin2A−
2
1
sin2B=0
Or
sin60
0
sin2A−cos60
0
sin2B=0
Or
cos60
0
sin2B−sin60
0
sin2A=0 ...(i)
Or
cos60
0
cos2A−sin60
0
sin2A=0 ... considering cos2A=sin2B
cos(60
0
−2A)=0
Or
60
0
−2A=90
0
,−90
0
Or
A=−15
0
,75
0
Now
cos2A=sin2B
Or
cos(−30
0
)=cos(
2
π
−2B)
Or
2B=120
0
Or
B=60
0
and
cos(150
0
)=cos(90
0
−2B)
Or
B=−30
0
Similarly from i we get
cos60
0
sin2B−sin60
0
sin2A=0
sin(2B−60
0
)=0 considering sin2A=cos2B.
Hence
2B−60
0
=0,180
0
Or
B=30
0
,120
0
Now
sin2A=cos2B
Or
sin2A=cos60
0
2A=30
0
A=15
0
And
sin2A=cos(120
0
)
sin2A=−
2
1
Or
2A=−30
0
Or
A=−15
0
.
Hence possible values of A {−15
0
,15
0
,75
0
}
Possible values of B {−30
0
,30
0
,60
0
,120
0
}
Now A and B are positive Acute angles
Hence Aϵ{15
0
,75
0
}
Bϵ{30
0
,60
0
}
However only 15
0
,30
0
satisfy the other equation.