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Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.

My work so far:

(I am replacing $\phi$ with the variable a for this)

$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given statement with 1.2 in it.)

$\sin^3 a + 3\sin a * \cos a (\sin a + \cos a) + \cos^3 a = 1.728$

$\sin^3 a + 3\sin a * \cos a (1.2) + \cos^3 a = 1.728$

$\sin^3 a + \cos^3 a = 1.728 - 3\sin a * \cos (a) *(1.2)$

Now I am stuck. What do I do next? Any hints?

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    $\begingroup$ Simplest way: $$\begin{split} \sin^3 a + \cos^3 a &= (\sin a + \cos a)(\sin^2 a + \cos^2 a - \sin a \cos a)\\ &= 1.2 \cdot(1 - \sin a \cos a) \end{split}$$ Rest you have in user201168's answer. $\endgroup$
    – Tacet
    Jan 5, 2015 at 23:09

6 Answers 6

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Hint: To finish your last step you need to find an expression for $\sin a\cos a$ in terms of $\sin a+\cos a$. I'll help you here :

$$ (\sin a+\cos a)^2=\color{green}{\sin^2 a}+2\color{#C00}{\sin a\cos a}+\color{green}{\cos^2 a}. $$

You're searching for the red part, note that the green part is equal to ...

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  • $\begingroup$ If I calculated correctly..0.936? $\endgroup$ Jan 5, 2015 at 23:31
  • $\begingroup$ @MathyPerson Correct. $\endgroup$
    – Workaholic
    Jan 6, 2015 at 7:10
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Note $$ x^3+y^3 = (x+y)\left(x^2-xy +y^2\right) $$ Now we can see that if we set $$ x =\sin a\\ y = \cos a. $$ Then we get $$ \sin^3a+\cos^3a = (\sin a +\cos a)\left(1-\sin a\cos a\right) $$ We can utilise $$ (\sin a+ \cos a)^2 = 1 +2\sin a\cos a $$ Combining all the above will yield $$ \sin^3a+\cos^3a = (\sin a +\cos a)\left(1-\frac{(\sin a+ \cos a)^2-1}{2}\right) $$

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Squaring $\sin a + \cos a = b$, $b^2 =\sin^2a+2\sin a \cos a + \cos^2 a = 1+2\sin a \cos a $, so $\sin a \cos a =(b^2-1)/2 $.

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Hint. Square both sides of your given equation. Use the fact that $\sin^2\phi+\cos^2\phi=1$ to find the value of $\sin\phi\cos\phi$. Together with the work you have done already, this solves the problem.

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i see that you already have several hints pointing you towards an answer. i will try a different method, perhaps a lengthier one. we will use the addition formula $$\sin (a+b) = \sin a \cos b + \sin b \cos a, \cos (a+b) = \cos a \cos b - \sin a \sin b$$ for $\sin()$ and $\cos().$

from $$\cos(t -45^\circ) = \cos 45^\circ \cos t + \sin 45^\circ \sin 45 = {1.2 \over \sqrt 2} = \cos a$$

we can get two things: (i) $t = a + 45^\circ$ and (ii) $\sin^2 a = 1 - \cos^2 a = 0.28$

now, we can evaluate $$\sqrt 2 \cos t = \sqrt 2 \cos(a+45^\circ) = \cos a - \sin a \text{ and } \sqrt 2 \sin t = \sqrt 2 \sin(a+45^\circ) = \cos a + \sin a$$

cubing and adding these you get $$ 2\sqrt 2(cos^3 t + \sin^3 t) = 2\cos a(\cos^2 a + 3 \sin^2 a) = 2*{1.2 \over \sqrt 2}(1 + 2*0.28)$$ so finally you get

$$ cos^3 t + \sin^3 t = 0.6*1.56 = 0.936$$

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You could use this $$\eqalign{ & {(\sin a + cosa)^3} = \sin {a^3} + cos{a^3} + 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr & \Rightarrow {\sin ^3}a + co{s^3}a = {(\sin a + cosa)^3} - 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr & and \cr & {(\sin a + cosa)^2} = 1 + 2\sin a \cdot cosa \cr & \Rightarrow \sin a \cdot cosa = {{{{(\sin a + cosa)}^2} - 1} \over 2} \cr & then : \sin a + cosa = k \cr & \Rightarrow {\sin ^3}a + co{s^3}a = {k^3} - {3 \over 2}k({k^2} - 1) = - {k \over 2}({k^2} - 3) \cr} $$

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