If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?

Question

If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?

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All I know is that $\sin^{3}a+\cos^{3}a$ is equal to $$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.

Answer 1

Hint:$$\left( \sin a+\cos a \right) \left( \sin ^{ 2 } a-\sin a\cos a+\cos ^{ 2 } a \right) =\\ =\left( \sin a+\cos a \right) \left( { \left( \sin a+\cos a \right) }^{ 2 }-\frac { 3 }{ 2 } \left( { \left( \sin { a } +\cos { a } \right) }^{ 2 }-1 \right) \right) $$

Answer 2

Hint: if $\sin\alpha+\cos\alpha=1.2$ then $$ (\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+2\sin\alpha\cos\alpha=1.44. $$

Answer 3

HINT: You know what $\sin^2\alpha+\cos^2\alpha$ is, so the problem boils down to sorting out $\sin\alpha\cos\alpha$. But

$$(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\;,$$

and you can solve this for $\sin\alpha\cos\alpha$.

Answer 4

$$\sin^3 \alpha+\cos^3 \alpha=(\sin \alpha+\cos \alpha)(\sin^2 \alpha-\sin \alpha\cos \alpha+\cos^2 \alpha)$$ $$1.2^3=\sin^3\alpha+\cos^3 \alpha+3\sin \alpha\cos \alpha(\sin \alpha+\cos \alpha)$$ It follows $$\sin \alpha\cos \alpha=0.22$$ and because $$\sin \alpha+\cos \alpha=1.2$$ one has sum and product of two unknowns so its are solution of the quadratic equation $$X^2-1.2X+0.22=0$$ with positive discriminant $0.14$.

Answer 5

${x^3+y^3 \over x+y} = x^2+y^2-xy$.

$(x+y)^2 = x^2+y^2+ 2 xy$, and so

${x^3+y^3 \over x+y} = x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2))$, from which we get

$x^3+y^3 = (x+y) (x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2)))$.

Since $x+y = 1.2, x^2+y^2 = 1$ we get $x^3+y^3 = (1.2)(1-{1 \over 2}((1.2^2-1)) = {117 \over 125}$.