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Question 1184287: The line joining A(bCosx, bSinx) and B(aCosy, aSiny), where a and b have different values is produced to the point M(x, y) so that AM:MB = b:a then find the value of uCos((x+y/2)) + vSin((x+y)/2)) and uCos(x+y) + vSin(x+y)?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let A = (b cos x, b sin x) and B = (a cos y, a sin y).
Let M = (u, v) be the point such that AM:MB = b:a.
Using the section formula for the coordinates of M, we have:
$$u = \frac{b(a \cos y) + a(b \cos x)}{b+a} = \frac{ab(\cos y + \cos x)}{a+b}$$
$$v = \frac{b(a \sin y) + a(b \sin x)}{b+a} = \frac{ab(\sin y + \sin x)}{a+b}$$
We are asked to find the values of:
1. $u \cos\left(\frac{x+y}{2}\right) + v \sin\left(\frac{x+y}{2}\right)$
2. $u \cos(x+y) + v \sin(x+y)$
1. $u \cos\left(\frac{x+y}{2}\right) + v \sin\left(\frac{x+y}{2}\right) = \frac{ab}{a+b} \left[ (\cos y + \cos x) \cos\left(\frac{x+y}{2}\right) + (\sin y + \sin x) \sin\left(\frac{x+y}{2}\right) \right]$
Using the sum-to-product formulas:
$\cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$
$\sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$
Substituting these into the expression:
$= \frac{ab}{a+b} \left[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right) + 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \sin\left(\frac{x+y}{2}\right) \right]$
$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \left[ \cos^2\left(\frac{x+y}{2}\right) + \sin^2\left(\frac{x+y}{2}\right) \right]$
$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right)$
2. $u \cos(x+y) + v \sin(x+y) = \frac{ab}{a+b} \left[ (\cos y + \cos x) \cos(x+y) + (\sin y + \sin x) \sin(x+y) \right]$
$= \frac{ab}{a+b} \left[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \cos(x+y) + 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \sin(x+y) \right]$
$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \left[ \cos\left(\frac{x+y}{2}\right) \cos(x+y) + \sin\left(\frac{x+y}{2}\right) \sin(x+y) \right]$
$= \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \cos\left(x+y - \frac{x+y}{2}\right) = \frac{2ab}{a+b} \cos\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right)$
Final Answer: The final answer is $\boxed{2ab/(a+b)cos((x-y)/2)}$
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