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If $$\sin(x) - \cos(x) = \frac{1}{3}$$ then determine $$\sin(x)\cos(x)$$

I know that the expected solution is squaring both sides of equation and solving it this way: \begin{gather} \sin^2(x)+\cos^2(x)= 1 \\[4px] (\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px] \sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px] -2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px] 2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px] 2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px] 2\sin(x)\cos(x)=\frac{8}{9} \\[4px] \sin(x)\cos(x)=\frac{4}{9} \end{gather} But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.

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  • $\begingroup$ I probably should have been more precise. I haven't learnt calculus yet if that makes any sense anyway. I'm on Algebra 2 level. $\endgroup$ – Jak Mar 13 at 22:30
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You can notice that $$ \sin x-\cos x=\sqrt{2}\sin(x-\pi/4) $$ Set $y=x-\pi/4$; then $\sin y=\frac{1}{3\sqrt{2}}$ and \begin{align} \sin x\cos x &=\sin(y+\pi/4)\cos(y+\pi/4) \\[4px] &=\left(\frac{1}{\sqrt{2}}\cos y+\frac{1}{\sqrt{2}}\sin y\right) \left(\frac{1}{\sqrt{2}}\cos y-\frac{1}{\sqrt{2}}\sin y\right) \\[4px] &=\frac{1}{2}(\cos^2y-\sin^2y) \\[4px] &=\frac{1}{2}(1-2\sin^2y) \\[4px] &=\frac{1}{2}\left(1-\frac{2}{(3\sqrt{2})^2}\right) \end{align}

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Note that $$\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = -\frac{1}{3\sqrt{2}}.$$ Now, using a double-angle formula: $$\cos(2x + \pi/2) = 2 \cos^2(x + \pi/4) - 1 = 2 \left( -\frac{1}{3\sqrt{2}}\right)^2 - 1 = -\frac{8}{9}.$$ On the other hand, $$\cos(2x + \pi/2) = \cos(2x) \cos(\pi/2) - \sin(2x) \sin(\pi/2) = -\sin(2x) = -2 \sin x \cos x.$$ Equating the two, $-2 \sin x \cos x = -\frac{8}{9}$.


In general, this method of using a sum-of-angles formula in reverse is very often useful in analyzing expressions of the form $A \cos x + B \sin x$ when $A$ and $B$ are constants, and is a nice tool to keep available in your toolbox.

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To save typing, let $s = \sin x$ and $c = \cos x$. Then we have two equations in two variables: $$ \begin{align} s^2 + c^2 &= 1\\ s - c &= \frac13. \end{align} $$ A widely applicable approach in such a situation is to solve one equation for one of the variables, and use that result to substitute into the other. Here, the second equation can easily be solved for $s$, yielding $s = c + \frac13$. Substituting for $s$ in the first equation gives a quadratic equation in $c$: $$ 2c^2 + \frac23 c - \frac89 = 0. $$ The roots of this quadratic are $$ c = \frac{-1+\sqrt{17}}{6}, \frac{-1-\sqrt{17}}{6}. $$ Since $s = c + \frac13$, the corresponding values for $s$ are $$ s = \frac{1+\sqrt{17}}{6}, \frac{1-\sqrt{17}}{6}. $$ Multiplying either pair of corresponding values gives the result $sc = \frac49$.

This method is much more laborious than the original solution, but requires no cleverness or flash of insight.

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You can use $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac {1-t^2}{1+t^2}$$

where $t=\tan (x/2)$, but you just have to know that some value of $t$ works.

Then you get a quadratic to solve for $t$. And once you know $t$ you can solve the problem.

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    $\begingroup$ This is a mechanical way of turning expressions involving sines and cosines into rational functions. See Weierstrass substitution (though the wikipedia article majors on calculus, this is really a parametrisation of the unit circle in the plane, and is useful for calculus, but not dependent on it - it works geometrically) $\endgroup$ – Mark Bennet Mar 13 at 22:40

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