# If $\sin^3(\theta)+\cos^3(\theta) = \frac{11}{16}$, find the exact value of $\sin(\theta) + \cos(\theta)$

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.

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

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?

• 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)? Commented May 11, 2021 at 19:28
• @johnwickww2312 What is $s^2 + c^2$? Commented May 11, 2021 at 21:22

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 $$\sin\theta + \cos\theta = 0.5$$.