# Solving a cubic equation involving trigonometric functions

Question: If $$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$ Then $$\sin^3(x) - \cos^3(x) = ?$$

I have turned first equation into a quadratic so I got $$\sin(x) = \frac{\sqrt{3}\mp\sqrt{5}}4$$ and $$\cos(x) = \frac{-\sqrt{3}\mp\sqrt{5}}4$$ But stuck here. I don't know what should I do, please help

Hint: We want $$(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x).$$

You should be able to find $\sin x\cos x$ by squaring $\sin x-\cos x$, which you know.

Remarks: $1$. We used the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.

$2$. I have not checked your calculations. However, from expressions for $\sin x$ and $\cos x$, you can compute $\sin^3 x-\cos^3 x$ by cubing. More painful, but doable. In the finding of $\sin x$, squaring was involved, so one has to be careful to use only the values of $\sin x,\cos x$ that satisfy the original equation.

• @satishramanathan: I can look at the question, but probably not until tomorrow, it is getting late here. – André Nicolas Apr 30 '15 at 6:17
• Nicolas, No problem sir. Thanks – Satish Ramanathan Apr 30 '15 at 6:18

$$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$ $$\implies \sin^2(x) + \cos^2(x)-2\sin x \cos x = \dfrac 34$$ $$\implies \sin x\cos x=\dfrac 18$$ (using $\sin^2x+\cos^2x=1$)

Now, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,$$\sin^3(x)-\cos^3x = (\sin x-\cos x)(\sin^2x+\cos^2x+\sin x\cos x)$$

$$=\dfrac {\sqrt 3}{2}\times \left (1+\dfrac 18\right)$$ $$=\dfrac {9\sqrt 3}{16}$$