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The equation is $$\sin^3(\theta)+\cos^3(\theta) = \frac{11}{16}$$ and it wants me to find the exact value of $\sin(\theta) + \cos(\theta)$.

I started at first trying to use Pythagorean identities, but those only work for squared trigs. I also tried to expand/use foil, but I'm stuck; not sure if this method is even the right one to use.

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    $\begingroup$ Can you factor a sum of two cubes? $\endgroup$
    – GEdgar
    Commented May 11, 2021 at 19:15
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    $\begingroup$ Could you explain what that is/how to do it? $\endgroup$ Commented May 11, 2021 at 19:16

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Hint: let $s = \sin \theta, c = \cos \theta$.

What you want to find is $s+c$.

You're given $s^3 + c^3$

Now $s^3 + c^3 = (s+c)(s^2 + c^2 - sc)$

And $2sc = (s+c)^2 - (s^2 + c^2)$

Can you finish?

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  • $\begingroup$ thanks for the help, assuming I had no errors I got cos^2(theta) and set it equal to 11/16, from here this will give me the value of theta and I then input it into the sin(theta) + cos(theta)? $\endgroup$ Commented May 11, 2021 at 19:28
  • $\begingroup$ @johnwickww2312 What is $s^2 + c^2$? $\endgroup$
    – Deepak
    Commented May 11, 2021 at 21:22
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For brevity, let $s = \sin\theta$ and $c = \cos\theta$.

We are given $s^3 + c^3 = \frac{11}{16}$, and know that $s^2 + c^2 = 1$ from Pythagoras. But what we ultimately want is $x = s + c$, so let's try to frame the equations in terms of $x$. First, let's evaluate some powers of $x$.

$$x^2 = (s + c)^2 = s^2 + 2cs + c^2 = 2cs + 1$$

From this, we get $cs = \frac{x^2 - 1}{2}$.

$$x^3 = (s + c)^3$$ $$x^3 = s^3 + 3cs^2 + 3c^2s + c^3$$ $$x^3 = (s^3 + c^3) + 3cs(s + c)$$ $$x^3 = \frac{11}{16} + 3(\frac{x^2 - 1}{2})(x)$$

We now have a cubic equation in terms of $x$ alone. Rearranging into standard descending-power order (and multiplying by 16 to eliminate fractions) gives:

$$8x^3 - 24x + 11 = 0$$

By the Rational Root Theorem, if $x$ is rational, then its numerator is a factor of 11 ($\pm\{1, 11\}$) and its denominator is a factor of 8 ($\pm\{1, 2, 4, 8 \}$). Trying all the possibilities gives $x = \frac{1}{2}$ as the only rational root. So the cubic has $2x - 1$ as a factor. After long division, the equation reduces to:

$$4x^2 + 2x - 11$$

Applying the quadratic formula gives $x = \frac{-2 \pm \sqrt{180}}{8} = \frac{-2 \pm 6 \sqrt{5}}{8} = \frac{-1 \pm 3 \sqrt{5}}{4}$. This gives us three potential solutions:

$$x = \frac{1}{2} = 0.5$$ $$x = \frac{-1 + 3 \sqrt{5}}{4} \approx 1.427051$$ $$x = \frac{-1 - 3 \sqrt{5}}{4} \approx -1.927051$$

To choose between the three possible options, use the identity

$$A\sin\theta + B\cos\theta = \sqrt{A^2 + B^2} \sin(\theta + \arctan(\frac{B}{A}))$$

with $A=B=1$, so:

$$\sin\theta + \cos\theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$$

The right-hand side of this equation is clearly bounded to the interval $[-\sqrt{2}, \sqrt{2}] \approx [-1.414214, 1.414214]$, ruling out the two irrational roots as out of range. Therefore, the only valid solution must be:

$$\boxed{\sin\theta + \cos\theta = \frac{1}{2}}$$

In case you're curious what $\theta$ is, I plugged the equation into a numeric solver and got two possible solutions (between $0$ and $2\pi$).

$$\theta \approx 1.994827 \approx 114.3°$$ $$\theta \approx 5.859154 \approx 335.7°$$

Note that these are complementary angles.

You can plug these numbers into a calculator to confirm that either one gives $\sin\theta + \cos\theta = 0.5$.

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