Trigonometry Examples

${(\frac{1}{4})}^{2} + y^{2} = 1$

Step 1

Simplify each term.

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Step 1.1

Apply the product rule to $\frac{1}{4}$ .

$\frac{1^{2}}{4^{2}} + y^{2} = 1$

Step 1.2

One to any power is one.

$\frac{1}{4^{2}} + y^{2} = 1$

Step 1.3

Raise $4$ to the power of $2$ .

$\frac{1}{16} + y^{2} = 1$

Step 2

Move all terms not containing $y$ to the right side of the equation.

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Step 2.1

Subtract $\frac{1}{16}$ from both sides of the equation.

$y^{2} = 1 - \frac{1}{16}$

Step 2.2

Write $1$ as a fraction with a common denominator.

$y^{2} = \frac{16}{16} - \frac{1}{16}$

Step 2.3

Combine the numerators over the common denominator.

$y^{2} = \frac{16 - 1}{16}$

Step 2.4

Subtract $1$ from $16$ .

$y^{2} = \frac{15}{16}$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{15}{16}}$

Step 4

Simplify $\pm \sqrt{\frac{15}{16}}$ .

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Step 4.1

Rewrite $\sqrt{\frac{15}{16}}$ as $\frac{\sqrt{15}}{\sqrt{16}}$ .

$y = \pm \frac{\sqrt{15}}{\sqrt{16}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Rewrite $16$ as $4^{2}$ .

$y = \pm \frac{\sqrt{15}}{\sqrt{4^{2}}}$

Step 4.2.2

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm \frac{\sqrt{15}}{4}$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{15}}{4}$

Step 5.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{\sqrt{15}}{4}$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{\sqrt{15}}{4}, - \frac{\sqrt{15}}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$y = \frac{\sqrt{15}}{4}, - \frac{\sqrt{15}}{4}$

Decimal Form:

$y = 0.96824583 \dots, - 0.96824583 \dots$