Question

$\tan^6 20^\circ - 33 \tan^4 20^\circ + 27 \tan^2 20^\circ$

$\sin^3 10^\circ + \sin^3 50^\circ - \sin^3 70^\circ$

Answer 1

The value of $\tan^6 20^\circ - 33 \tan^4 20^\circ + 27 \tan^2 20^\circ$ is $3$.

The value of $\sin^3 10^\circ + \sin^3 50^\circ - \sin^3 70^\circ$ is $-\frac{3}{8}$.

Answer 2

To solve the given expressions, we will address each one separately.

Part 1: $\tan^6 20^\circ - 33 \tan^4 20^\circ + 27 \tan^2 20^\circ$

Let $\theta = 20^\circ$. We know that $3\theta = 60^\circ$. We use the triple angle identity for tangent:

$$\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$$

Substitute $\theta = 20^\circ$:

$$\tan 60^\circ = \frac{3 \tan 20^\circ - \tan^3 20^\circ}{1 - 3 \tan^2 20^\circ}$$

We know $\tan 60^\circ = \sqrt{3}$. Let $t = \tan 20^\circ$.

$$\sqrt{3} = \frac{3t - t^3}{1 - 3t^2}$$

Cross-multiply:

$$\sqrt{3}(1 - 3t^2) = 3t - t^3$$ $$\sqrt{3} - 3\sqrt{3}t^2 = 3t - t^3$$

Rearrange the terms to isolate terms with $\sqrt{3}$:

$$t^3 - 3t = 3\sqrt{3}t^2 - \sqrt{3}$$

Factor out $t$ on the left and $\sqrt{3}$ on the right:

$$t(t^2 - 3) = \sqrt{3}(3t^2 - 1)$$

Square both sides of the equation:

$$[t(t^2 - 3)]^2 = [\sqrt{3}(3t^2 - 1)]^2$$ $$t^2(t^2 - 3)^2 = 3(3t^2 - 1)^2$$

Expand both sides:

$$t^2(t^4 - 6t^2 + 9) = 3(9t^4 - 6t^2 + 1)$$ $$t^6 - 6t^4 + 9t^2 = 27t^4 - 18t^2 + 3$$

Move all terms to one side to match the given expression:

$$t^6 - 6t^4 - 27t^4 + 9t^2 + 18t^2 - 3 = 0$$ $$t^6 - 33t^4 + 27t^2 - 3 = 0$$

Therefore,

$$\tan^6 20^\circ - 33 \tan^4 20^\circ + 27 \tan^2 20^\circ = 3$$

Part 2: $\sin^3 10^\circ + \sin^3 50^\circ - \sin^3 70^\circ$

Let $x = \sin 10^\circ$, $y = \sin 50^\circ$, and $z = -\sin 70^\circ$. We need to evaluate $x^3 + y^3 + z^3$. We know that if $x+y+z=0$, then $x^3+y^3+z^3 = 3xyz$. Let's check if $x+y+z = 0$:

$$\sin 10^\circ + \sin 50^\circ - \sin 70^\circ$$

Using the sum-to-product identity $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$:

$$\sin 10^\circ + \sin 50^\circ = 2 \sin\left(\frac{10^\circ+50^\circ}{2}\right) \cos\left(\frac{50^\circ-10^\circ}{2}\right)$$ $$= 2 \sin 30^\circ \cos 20^\circ$$ $$= 2 \times \frac{1}{2} \times \cos 20^\circ = \cos 20^\circ$$

Now substitute this back into the sum:

$$\cos 20^\circ - \sin 70^\circ$$

Since $\sin 70^\circ = \sin (90^\circ - 20^\circ) = \cos 20^\circ$:

$$\cos 20^\circ - \cos 20^\circ = 0$$

So, $x+y+z=0$ holds true. Therefore, $\sin^3 10^\circ + \sin^3 50^\circ - \sin^3 70^\circ = 3 (\sin 10^\circ)(\sin 50^\circ)(-\sin 70^\circ)$.

$$= -3 \sin 10^\circ \sin 50^\circ \sin 70^\circ$$

We use the product identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$. Here, $\theta = 10^\circ$. So, $60^\circ - \theta = 50^\circ$ and $60^\circ + \theta = 70^\circ$.

$$\sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4} \sin (3 \times 10^\circ)$$ $$= \frac{1}{4} \sin 30^\circ$$ $$= \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$

Substitute this value back into the expression:

$$-3 \times \frac{1}{8} = -\frac{3}{8}$$

Explanation of the solution:

1. For the first expression:

   • Let $t = \tan 20^\circ$. Use the triple angle identity $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$ with $\theta = 20^\circ$.
   • This gives $\tan 60^\circ = \sqrt{3} = \frac{3t - t^3}{1 - 3t^2}$.
   • Rearrange the equation to $t(t^2 - 3) = \sqrt{3}(3t^2 - 1)$.
   • Square both sides: $t^2(t^2 - 3)^2 = 3(3t^2 - 1)^2$.
   • Expand and simplify: $t^2(t^4 - 6t^2 + 9) = 3(9t^4 - 6t^2 + 1) \implies t^6 - 6t^4 + 9t^2 = 27t^4 - 18t^2 + 3$.
   • Collect terms: $t^6 - 33t^4 + 27t^2 = 3$.

2. For the second expression:

   • Let $x = \sin 10^\circ$, $y = \sin 50^\circ$, $z = -\sin 70^\circ$.
   • Check if $x+y+z=0$. $\sin 10^\circ + \sin 50^\circ - \sin 70^\circ = 2\sin 30^\circ \cos 20^\circ - \sin 70^\circ = \cos 20^\circ - \cos 20^\circ = 0$.
   • Since $x+y+z=0$, then $x^3+y^3+z^3 = 3xyz$.
   • So, $\sin^3 10^\circ + \sin^3 50^\circ - \sin^3 70^\circ = -3 \sin 10^\circ \sin 50^\circ \sin 70^\circ$.
   • Use the product identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$ with $\theta = 10^\circ$.
   • $\sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4} \sin (3 \times 10^\circ) = \frac{1}{4} \sin 30^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
   • Substitute back: $-3 \times \frac{1}{8} = -\frac{3}{8}$.