SOLUTION: A*sin(X)+B*sin(X+U)=C*sin(X+W), with C=sqrt(A^2+B^2+2*A*B*cos(U))
and tan(W)=(B*sin(U))/(A+B*cos(U)).
What are the intermediate steps leading to this result?
Algebra ->
Trigonometry-basics
-> SOLUTION: A*sin(X)+B*sin(X+U)=C*sin(X+W), with C=sqrt(A^2+B^2+2*A*B*cos(U))
and tan(W)=(B*sin(U))/(A+B*cos(U)).
What are the intermediate steps leading to this result?
Log On
Question 1087083: A*sin(X)+B*sin(X+U)=C*sin(X+W), with C=sqrt(A^2+B^2+2*A*B*cos(U))
and tan(W)=(B*sin(U))/(A+B*cos(U)).
What are the intermediate steps leading to this result? Answer by Edwin McCravy(20086) (Show Source):
You can put this solution on YOUR website! A*sin(X)+B*sin(X+U)=C*sin(X+W), with C=sqrt(A^2+B^2+2*A*B*cos(U))
and tan(W)=(B*sin(U))/(A+B*cos(U)).
We are given that tan(W)=(B*sin(U))/(A+B*cos(U)).
Since and
,
we can draw a right triangle with B*sin(U) as the
opposite side of angle W and A+B*cos(U) as the
adjacent side of W:
We calculate the hypotenuse:
Notice that this result is the exact value that was given for C. So
Let's start with right side of what we have to prove:
Since and ,
we use the right triangle above to substitute for cos(W) and sin(W)
Edwin
