5 problems found
Solution:
Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]
Solution: \begin{align*} && \sin(x+y) - \sin(x-y) &= \sin x \cos y + \cos x \sin y - (\sin x \cos y - \cos x \sin y )\\ &&&= 2 \cos x \sin y \\ \\ && A &= x+y \\ && B &= x - y \\ \Rightarrow && x = \frac12(A+B) &\quad y = \frac12(A-B) \\ \Rightarrow && \sin A - \sin B &= 2 \cos \tfrac12(A+B) \sin \tfrac12(A-B) \\ \\ && \cos (x+y) - \cos (x-y) &= \cos x \cos y - \sin x \sin y -(\cos x \cos y + \sin x \sin y ) \\ &&&= -2 \sin x \sin y \\ \Rightarrow && \cos A - \cos B &= - 2 \sin \tfrac12 (A+B) \sin \tfrac12 (A-B) \end{align*} \begin{align*} && \text{Gradient of }PQ &= \frac{b \sin q - b \sin p}{a \cos q - a \cos p } \\ && \text{Gradient of }SR &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ PQ \parallel SR \Rightarrow && \frac{b \sin q - b \sin p}{a \cos q - a \cos p } &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\ \Rightarrow && (\sin q - \sin p)(\cos s - \cos r) &= (\sin s - \sin r)(\cos q - \cos r) \\ \Rightarrow && -4 \cos \tfrac12(p+q) \sin\tfrac12(q-p) \sin \tfrac12(s+r) \sin \tfrac12(s-r) &= -4 \cos \tfrac12(s+r) \sin \tfrac12(s-r) \sin \tfrac12 (p+q) \sin\tfrac12 (q-p) \\ \Rightarrow && 0 &= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \left ( \cos \tfrac12(p+q)\sin \tfrac12(s+r) - \sin \tfrac12 (p+q)\cos \tfrac12(s+r) \right) \\ &&&= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \sin \left ( \frac12 (s+r -(p+q))\right) \end{align*} Since \(s \neq r\) and \(p \neq q\) (neither line vertical) we must have \(\frac12 (s+r -(p+q)) = n \pi \Rightarrow s+r - p - q = 0, 2\pi, 4\pi, \cdots\) but given the range constraints it must be \(2 \pi\)
The point \(P\) has coordinates \((x,y)\) with respect to the origin \(O\). By writing \(x=r\cos\theta\) and \(y=r\sin\theta\), or otherwise, show that, if the line \(OP\) is rotated by \(60^\circ\) clockwise about \(O\), the new \(y\)-coordinate of \(P\) is \(\frac12(y-\sqrt3\,x)\). What is the new \(y\)-coordinate in the case of an anti-clockwise rotation by \(60^\circ\,\)? An equilateral triangle \(OBC\) has vertices at \(O\), \((1,0)\) and \((\frac12,\frac12 \sqrt3)\), respectively. The point \(P\) has coordinates \((x,y)\). The perpendicular distance from \(P\) to the line through \(C\) and \(O\) is \(h_1\); the perpendicular distance from \(P\) to the line through \(O\) and \(B\) is \(h_2\); and the perpendicular distance from \(P\) to the line through \(B\) and \(C\) is \(h_3\). Show that \(h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert\) and find expressions for \(h_2\) and \(h_3\). Show that \(h_1+h_2+h_3=\frac12 \sqrt3\) if and only if \(P\) lies on or in the triangle \(OBC\).
A curve has the equation \(y=\f(x)\), where \[ \f(x) = \cos \Big( 2x+ \frac \pi 3\Big) + \sin \Big ( \frac{3x}2 - \frac \pi 4\Big). \]
Solution: \begin{align*} && f(x) &= \cos \left( 2x+ \frac \pi 3\right) + \sin \left ( \frac{3x}2 - \frac\pi 4\right) \\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{\pi}{2} - \left ( \frac{3x}2 - \frac\pi 4\right) \right)\\ &&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{3\pi}{4} - \frac{3x}2 \right)\\ &&&= 2 \cos \left (\frac{2x+ \frac \pi 3+\frac{3\pi}{4} - \frac{3x}2}{2} \right) \cos \left ( \frac{\left (2x+ \frac \pi 3 \right) - \left (\frac{3\pi}{4} - \frac{3x}2 \right)}{2} \right)\\ &&&= 2 \cos \left (\frac{\frac{x}{2}+ \frac {13\pi}{12}}{2} \right) \cos \left ( \frac{\frac{7x}{2}- \frac {5\pi}{12}}{2} \right)\\ &&&= 2 \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right)\\ \end{align*}
Given that \(\cos A\), \(\cos B\) and \(\beta\) are non-zero, show that the equation \[ \alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B) \] reduces to the form \[ (\tan A-m)(\tan B-n)=0\,, \] where \(m\) and \(n\) are independent of \(A\) and \(B\), if and only if \(\alpha^2=\beta^2+\gamma^2\). Determine all values of \(x\), in the range \(0\le x <2\pi\), for which:
Solution: \begin{align*} && \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\ \Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\ \Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\ \Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\ \Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\ \end{align*} Which has the desired form iff \(\beta^2+\gamma^2 = \alpha^2\).