Trigonometry Examples

$(3, - \sqrt{7})$

Step 1

To find the $\cos (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(3, - \sqrt{7})$ , draw the triangle between the three points $(0, 0)$ , $(3, 0)$ , and $(3, - \sqrt{7})$ .

Opposite : $- \sqrt{7}$

Adjacent : $3$

Step 2

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

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Step 2.1

Simplify the expression.

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Step 2.1.1

Raise $3$ to the power of $2$ .

$\sqrt{9 + {(- \sqrt{7})}^{2}}$

Step 2.1.2

Apply the product rule to $- \sqrt{7}$ .

$\sqrt{9 + {(- 1)}^{2} {\sqrt{7}}^{2}}$

Step 2.1.3

Raise $- 1$ to the power of $2$ .

$\sqrt{9 + 1 {\sqrt{7}}^{2}}$

Step 2.1.4

Multiply ${\sqrt{7}}^{2}$ by $1$ .

$\sqrt{9 + {\sqrt{7}}^{2}}$

Step 2.2

Rewrite ${\sqrt{7}}^{2}$ as $7$ .

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

$\sqrt{9 + {(7^{\frac{1}{2}})}^{2}}$

Step 2.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{9 + 7^{\frac{1}{2} \cdot 2}}$

Step 2.2.3

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{9 + 7^{\frac{2}{2}}}$

Step 2.2.4

Cancel the common factor of $2$ .

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Step 2.2.4.1

Cancel the common factor.

$\sqrt{9 + 7^{\frac{2}{2}}}$

Step 2.2.4.2

Rewrite the expression.

$\sqrt{9 + 7^{1}}$

Step 2.2.5

Evaluate the exponent.

$\sqrt{9 + 7}$

Step 2.3

Simplify the expression.

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Step 2.3.1

Add $9$ and $7$ .

$\sqrt{16}$

Step 2.3.2

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

$\sqrt{4^{2}}$

Step 2.3.3

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

$4$

Step 3

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{3}{4}$ .

$\frac{3}{4}$

Step 4

Approximate the result.

$\cos (θ) = \frac{3}{4} \approx 0.75$