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$\sin x- \cos x= \frac {1}{2}$; $\sin^3 x- \cos^3 x =$ ?

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    I have edited the question with mathjax. You should include your attempts for a favourable response. – 2017-01-21
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    $\sin 2x=\frac34$ on the other hand $\sin^3 x- \cos^3 x=(\sin x-\cos x)^3+\frac{3}{2}\sin 2x(\sin x-\cos x)=\frac {11}{16}$ – 2017-01-21

4 Answers 4

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Hint:

Note the following: $$\sin^3(x)-\cos^3(x)=(\sin(x)-\cos(x))(1+\sin(x)\cos(x))$$ And $$(\sin(x)-\cos(x))^2=\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x)=1-2\sin(x)\cos(x)=\frac{1}{4}$$

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$$\sin x -\cos x=\frac12$$ $$\sin^2x-2\sin x\cos x+\cos^2x=\frac14$$ $$\sin x\cos x=?$$

Also $$\sin^3 x- \cos^3 x =(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)=?$$

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sin x - cos x = $\frac12$

Squaring both sides,

$\sin^2 x + \cos^2 x - 2 \sin x \cos x = \frac 14$

$1 - 2 \sin x \cos x = \frac 14$

$- 2 \sin x \cos x = \frac 14 - 1$

$- 2 \sin x \cos x = - \frac 34$

$\sin x \cos x = \frac 38$

Now $\sin^3 x - \cos^3 x$

= $(\sin x-\cos x)(\sin^2 x+\cos^2 x+\sin x \cos x)$

= $\left( \frac12 \right) \left(1 + \frac38\right)$

= $\left( \frac12 \right) \left( \frac{11}8\right)$

= $\left( \frac{11}{16}\right) $

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    Your answer easiest for me in the all answer, but it is not true answer. Thanks. – 2017-01-21
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    @K You made a mistake: $$-2\sin x\cos x=-\frac34\implies \sin x\cos x=\frac38$$ – 2017-01-21
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    @DonAntonio thank u. My main problem is I am used mobile to solve so some time typos. – 2017-01-21
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    Yes I realised your mistake and fixed it:) Your answer is best for me :) – 2017-01-21
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    Mine pleasure.. – 2017-01-21
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use that $$\sin(x)-\cos(x)=-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)$$

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    How do you proceed after that? You have $\sin(x-\frac{\pi}{4})=\frac{1}{2\sqrt{2}}$ – 2017-01-21
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    we have $$x-\frac{\pi}{4}=\arcsin\left(\frac{1}{2\sqrt{2}}\right)$$ – 2017-01-21
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    the result should be $$\frac{11}{16}$$ – 2017-01-21
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    Because $\sin x$ is not $1-1$, we cannot simply take the inverse of this function on both sides without losing generality. – 2017-01-21
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    do you actually compute every possible value of sin(x) and cos(x)? It's a bit long but certainly yields the result. – 2017-01-21
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    yes but i have not squared the given equation! – 2017-01-21
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    It doesn't make a difference. sin(x) = y, has multiple solutions for x without squaring. E.g: sin(0)=sin($2\pi$)=0. – 2017-01-21
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    The general solution of $sin(x)=t, 0\leq t\leq 1$ is $x = (-1)^n(2n+1)*\frac{\pi}{2}+sin^{-1}t$ where $sin^{-1}t$ denotes the value between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$ Good luck expanding that. – 2017-01-21