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Given $$\left(\;\sin(x-y)+\cos(x+2y)\sin y\;\right)^2=4\cos x\sin y\sin(x+y)$$ prove $$\tan x+\tan y=\frac{\tan y}{(\sqrt{2}\cos y-1)^2}$$

My attempt:

$$\left(\;\sin(x-y)+0.5\sin(x+3y)-0.5\sin x\;\right)^2=2\left(\;\sin(x+y)-\sin(x-y)\;\right)$$

I am not be able to prove last line. Help required.

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  • $\begingroup$ MathJax hint: if you use \left[ and \right] instead of \bigg the brackets autosize for what is inside. That way you don't have to keep up with changes. It works on all delimiters. You do get an error message when they are mismatched, though. $\endgroup$ – Ross Millikan Apr 26 '18 at 15:31
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Are known the trig identities \begin{align} &\sin(a\pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b),\tag1\\[4pt] &\cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1 = 1-2\sin^2(a),\tag2\\[4pt] &\cos(a) - \cos(b) = 2\sin\dfrac{a+b}2\sin\dfrac{b-a}2.\tag3\\ \end{align} Using $(1) - (3),$ one can reduce the arguments of the LHS of the issue condition, \begin{align} &\sin(x-y)+\cos(x+2y)\sin (y) = \sin(x)\cos(y)+\cos(x+2y)\sin (y)-\cos(x)\sin(y) = \\[4pt] &\sin(x)\cos(y)-2\sin(x+y)\sin^2(y) = (1-2\sin^2(y))\sin(x+y)- \sin(y)\cos(x) =\\[4pt] &\cos(2y)\sin(x+y)- \sin(y)\cos(x).\\[4pt] \end{align}

Then \begin{align} &(\cos(2y)\sin(x+y)- \sin(y)\cos(x))^2 = 4\sin(y)\cos(x)\sin(x+y),\\ &\cos^2(2y)\sin^2(x+y)-2(\cos(2y)+2)\sin(y)\cos(x)\sin(x+y)+\sin^2(y)\cos^2(x) = 0.\tag4\\ \end{align}

Due to the area of admissible values of the goal equality, $\cos(x)\not=0$ and $\cos(y)\not=0.$ So it is correct to divide the equation $(4)$ by $\cos^2(x)\cos^2(y).$

Taking in account $(1)-(2),$ this leads to the square equation in the form of \begin{align} &\cos^2(2y)(\tan(x)+\tan(y))^2-2(\cos(2y)+2)\tan(y)(\tan(x)+\tan(y))^2+\tan^2(y) = 0,\\[4pt] &D = (\cos(2y)+2)^2 - \cos^2(2y) = 4(\cos(2y)+1) = 8\cos^2(y),\\[4pt] &\tan(x)+\tan(y) = \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{\cos^2(2y)}\tan(y),\\[4pt] &\dfrac{\cos(2y)+2\pm2\sqrt2\ \cos(y)}{\cos^2(2y)} = \dfrac{2\cos^2(y)\pm2\sqrt2\ \cos(y) +1}{(2\cos^2(y)-1)^2} = \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{(2\cos^2(y)-1)^2}\\[4pt] &= \dfrac{\left(\sqrt2\cos(y)\pm1\right)^2}{\left(\sqrt2\cos(y)+1\right)^2\left(\sqrt2\cos(y)-1\right)^2} = \dfrac1{\left(\sqrt2\cos(y)\pm1\right)^2},\\[4pt] &\tan(x)+\tan(y) = \dfrac{\tan(y)}{\left(\sqrt2\cos(y)\pm1\right)^2}.\tag5\\[4pt] \end{align} In accordance with the formula $(5),$ the initial problem setting looks incorrect, with a minimal discrepancy in the answers.

Maybe something is missing in the conditions?

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You can approach this depending on your toolset. I prefer factoring expressions. Define $$ X := e^{ix}, \quad Y := e^{iy}, \quad A^- := \tan(x)+\tan(y)-\tan(y)/(1-\sqrt{2}\cos(y))^2, $$ $$ \quad A^+ := \tan(x)+\tan(y)-\tan(y)/(1+\sqrt{2}\cos(y))^2, $$ $$\quad C^- := 1 - \sqrt{2}Y(1+X) + XY^2, \quad C^+ := 1 + \sqrt{2}Y(1+X) + XY^2, $$ $$\quad D^- := 1 - \sqrt{2}Y(1-X) - XY^2, \quad D^+ := 1 + \sqrt{2}Y(1-X) - XY^2, $$ $$ B := (\sin(x-y)+\cos(x+2y)\sin(y))^2-4\cos(x)\sin(y)\sin(x+y),$$ where we used superscript ${^-}$ and ${^+}$ for $-\sqrt{2}$ and $+\sqrt{2},\;$ respectively. Using factoring we get that $$\;A^- = 2iC^-D^-/((1+X^2)(1-\sqrt{2}Y+Y^2)^2), \quad B = -C^-D^-C^+D^+ (1+Y^2)^2/(4XY^3)^2,\;$$ $$\;A^+ = 2iC^+D^+/((1+X^2)(1-\sqrt{2}Y+Y^2)^2).$$ If the numerator of $A^-$ is zero, then so is $B$. This proves $A^-=0$ implies $B=0.\;$ Similarly, if the numerator of $A^+$ is zero, then so is $B$. This proves $A^+=0$ implies $B=0.\;$ Finally, if $\;B=0\;$ then either $\;A^-=0\;$ or $\;A^+=0.$

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