If 3sinA+4cosA=5, find sinA
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3/5 Sin A + 4/5 Cos A = 1
we see that (3/5)² + (4/5)² = 1
let 3/5 = Sin B and 4/5 = Cos B
So Sin A Sin B + Cos A Cos B = 1 = Cos 0°
Cos (A - B) = cos 0°
A = B or A - B = 2 π
if A = B, Sin A = Sin B = 3/5.
or, if A - B = 2π, Sin A = Sin (2π+B) = Sin B = 3/5
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alternately:
4 Cos A = 5 - 3 Sin A
16 Cos² A = 25 + 9 Sin² A - 30 Sin A
16 - 16 Sin² A = 25 + 9 Sin² A - 30 Sin A
25 Sin² A - 30 Sin A + 9 = 0
Sin A = [ 30 +- √(900 - 900) ] / 50 = 3/5
we see that (3/5)² + (4/5)² = 1
let 3/5 = Sin B and 4/5 = Cos B
So Sin A Sin B + Cos A Cos B = 1 = Cos 0°
Cos (A - B) = cos 0°
A = B or A - B = 2 π
if A = B, Sin A = Sin B = 3/5.
or, if A - B = 2π, Sin A = Sin (2π+B) = Sin B = 3/5
=========================
alternately:
4 Cos A = 5 - 3 Sin A
16 Cos² A = 25 + 9 Sin² A - 30 Sin A
16 - 16 Sin² A = 25 + 9 Sin² A - 30 Sin A
25 Sin² A - 30 Sin A + 9 = 0
Sin A = [ 30 +- √(900 - 900) ] / 50 = 3/5
Anniieee:
Thanx a lot.. u helped me very much.. just one ques how did sinAsinB+ cosAcosB become cos(A-B)?
that is the formula for Cos (A - B).
Sin A Cos B + Cos A Sin B = Sin (A+B) is another formul
Kk.. thanx
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guys I don't know the answer bcoz I m also finding this question's answer..
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