10
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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 '15 at 23:09
12
$\begingroup$

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 ...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If I calculated correctly..0.936? $\endgroup$ – Mathy Person Jan 5 '15 at 23:31
  • $\begingroup$ @MathyPerson Correct. $\endgroup$ – Workaholic Jan 6 '15 at 7:10
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.