Trigonometry Examples

$\arccos (x) + \arccos (2 x) = \frac{π}{3}$

Step 1

If $a = b$ , then $\cos (a) = \cos (b)$ .

$\cos (\arccos (x) + \arccos (2 x)) = \cos (\frac{π}{3})$

Step 2

Expand the left side.

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Step 2.1

Apply the sum of angles identity $\cos (x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ .

$\cos (\arccos (x)) \cos (\arccos (2 x)) - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2

Simplify each term.

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Step 2.2.1

The functions cosine and arccosine are inverses.

$x \cos (\arccos (2 x)) - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.2

The functions cosine and arccosine are inverses.

$x (2 x) - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.3

Rewrite using the commutative property of multiplication.

$2 x \cdot x - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.4

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.4.1

Move $x$ .

$2 (x \cdot x) - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.4.2

Multiply $x$ by $x$ .

$2 x^{2} - \sin (\arccos (x)) \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.5

Draw a triangle in the plane with vertices $(x, \sqrt{1^{2} - x^{2}})$ , $(x, 0)$ , and the origin. Then $\arccos (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(x, \sqrt{1^{2} - x^{2}})$ . Therefore, $\sin (\arccos (x))$ is $\sqrt{1 - x^{2}}$ .

$2 x^{2} - \sqrt{1 - x^{2}} \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.6

Rewrite $1$ as $1^{2}$ .

$2 x^{2} - \sqrt{1^{2} - x^{2}} \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$2 x^{2} - \sqrt{(1 + x) (1 - x)} \sin (\arccos (2 x)) = \cos (\frac{π}{3})$

Step 2.2.8

Draw a triangle in the plane with vertices $(2 x, \sqrt{1^{2} - {(2 x)}^{2}})$ , $(2 x, 0)$ , and the origin. Then $\arccos (2 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(2 x, \sqrt{1^{2} - {(2 x)}^{2}})$ . Therefore, $\sin (\arccos (2 x))$ is $\sqrt{1 - {(2 x)}^{2}}$ .

$2 x^{2} - \sqrt{(1 + x) (1 - x)} \sqrt{1 - {(2 x)}^{2}} = \cos (\frac{π}{3})$

Step 2.2.9

Rewrite $1$ as $1^{2}$ .

$2 x^{2} - \sqrt{(1 + x) (1 - x)} \sqrt{1^{2} - {(2 x)}^{2}} = \cos (\frac{π}{3})$

Step 2.2.10

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 2 x$ .

$2 x^{2} - \sqrt{(1 + x) (1 - x)} \sqrt{(1 + 2 x) (1 - (2 x))} = \cos (\frac{π}{3})$

Step 2.2.11

Multiply $2$ by $- 1$ .

$2 x^{2} - \sqrt{(1 + x) (1 - x)} \sqrt{(1 + 2 x) (1 - 2 x)} = \cos (\frac{π}{3})$

Step 2.2.12

Combine using the product rule for radicals.

$2 x^{2} - \sqrt{(1 + 2 x) (1 - 2 x) (1 + x) (1 - x)} = \cos (\frac{π}{3})$

Step 3

Expand the right side.

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Step 3.1

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$2 x^{2} - \sqrt{(1 + 2 x) (1 - 2 x) (1 + x) (1 - x)} = \frac{1}{2}$

Step 4

Solve the equation for $x$ .

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Step 4.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = - \frac{1}{2}, \frac{1}{2}$

Step 5

Exclude the solutions that do not make $\arccos (x) + \arccos (2 x) = \frac{π}{3}$ true.

$x = \frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}$

Decimal Form:

$x = 0.5$