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Help me out with a trig identities question?
If sin(A+B)=0.75 and sin(A-B)=0.43, then the value of sinBcosA, to the nearest hundredth, is_______
If you could walk me through it that'd be awesome. Quick Best Answer.
4 Answers
- Greg GLv 510 years agoFavorite Answer
From the first equation we get:
sin(A)cos(B) + Sin(B)cos(A) = 0.75
For the 2nd equation, sin(A-B) = Sin(A + -B) = sin(A)cos(-B) + sin(-B)cos(A) = 0.43
Recall that: sin(-x) = -sin(x) and cos(-x) = cos(x)
Our 2nd equation becomes:
sin(A)cos(B) - sin(B)cos(A) = 0.43
If we add our equations together we get:
2sin(A)cos(B) = 0.75 + 0.43
sin(A)cos(B) = 0.59
From our 2nd equation: 0.59 - sin(B)cos(A) = 0.43 ---> sin(B)cos(A) = 0.16
- tk2Lv 410 years ago
(1) sin(A+B) = sinAcosB + sinBcosA
(2) sin(A-B) = sinAcosB -sinBcosA
(1) - (2) = 2sinBcosA
0.75-0.53=2sinBcosA
sinBcosA = (0.75-0.53)/2
- s kLv 710 years ago
sin(x + y) = sin(x)cos(y) + cos(x)sin(y) = .75
sin(x - y) = sin(x)cos(y) - cos(x)sin(y) = .43
==>
sin(x)cos(y) + cos(x)sin(y) = .75
+ (-sin(x)cos(y) + cos(x)sin(y) = -.43)
=> 2cos(x)sin(y) = .32
=> cos(x)sin(y) = .16
- henry_yang67Lv 610 years ago
sin(A+B) = sinAcosB+ cosAsinB
sin(A-B) = sinAcosB-cosAsinB
sum them up, 2sinAcosB = 1.18
so sinAcosB = 0.59
