+0

0
149
2

sin(a+b)=0.86

sin(a-b)=0.34

What is sinbcosa?

(I did this question previously on the test by making two formulas, one for a and one for b and substituting the formulas into each other to figure out a and b separately, but now I don't remember how to do it (ps answer is 0.26)

Guest Mar 1, 2017
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#1
+84285
0

sin (a + b)   =  sinacosb + sinbcosa  = .86

sin (a - b)   =   sinacosb - sinbcosa  = .34      add these

2sinacosb  =  1.20          divide by 2

sinacosb  = .60

Sub this into either of the first two equations

.60 + sinbcosa  = .86     subtract .60 from each side

sinbcosa   = .26

CPhill  Mar 1, 2017
#2
+19089
+5

sin(a+b)=0.86

sin(a-b)=0.34

What is sinbcosa?

$$\small{ \begin{array}{|lrcll|} \hline (1) & \sin(a+b) &=& \sin(a)\cos(b)+\cos(a)\sin(b) \\ (2) & \sin(a-b) &=& \sin(a)\cos(b)-\cos(a)\sin(b) \\ \hline (1)-(2): & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)-[~\sin(a)\cos(b)-\cos(a)\sin(b)~] \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)- \sin(a)\cos(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)- \sin(a)\cos(b)+\cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\sin(b)\cos(a) \\ & \sin(b)\cos(a) &=& \frac12\cdot [\sin(a+b)-\sin(a-b)] \quad | \quad \sin(a+b)=0.86 \quad \sin(a-b)=0.34 \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.86-0.34) \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.52) \\ & \mathbf{\sin(b)\cos(a)} & \mathbf{=} & \mathbf{0.26} \\ \hline \end{array} }$$

heureka  Mar 1, 2017
edited by heureka  Mar 1, 2017

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