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)
+128400 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
+26367 sin(a+b)=0.86
sin(a-b)=0.34
What is sinbcosa?
\(\small{ \begin{array}{|lrcll|} \hline (1) & \sin(a+b) &=& \sin(a)\cos(b)+\cos(a)\sin(b) \\ (2) & \sin(a-b) &=& \sin(a)\cos(b)-\cos(a)\sin(b) \\ \hline (1)-(2): & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)-[~\sin(a)\cos(b)-\cos(a)\sin(b)~] \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)- \sin(a)\cos(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)- \sin(a)\cos(b)+\cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\sin(b)\cos(a) \\ & \sin(b)\cos(a) &=& \frac12\cdot [\sin(a+b)-\sin(a-b)] \quad | \quad \sin(a+b)=0.86 \quad \sin(a-b)=0.34 \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.86-0.34) \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.52) \\ & \mathbf{\sin(b)\cos(a)} & \mathbf{=} & \mathbf{0.26} \\ \hline \end{array} } \)
