Trigonometry Examples

$(1, \sqrt{2})$

Step 1

To find the $\sec (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(1, \sqrt{2})$ , draw the triangle between the three points $(0, 0)$ , $(1, 0)$ , and $(1, \sqrt{2})$ .

Opposite : $\sqrt{2}$

Adjacent : $1$

Step 2

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

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Step 2.1

One to any power is one.

$\sqrt{1 + {(\sqrt{2})}^{2}}$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

$\sqrt{1 + 2^{\frac{2}{2}}}$

Step 2.2.4

Cancel the common factor of $2$ .

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Step 2.2.4.1

Cancel the common factor.

$\sqrt{1 + 2^{\frac{2}{2}}}$

Step 2.2.4.2

Rewrite the expression.

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

Step 2.2.5

Evaluate the exponent.

$\sqrt{1 + 2}$

Step 2.3

Add $1$ and $2$ .

$\sqrt{3}$

Step 3

$\sec (θ) = \frac{Hypotenuse}{Adjacent}$ therefore $\sec (θ) = \frac{\sqrt{3}}{1}$ .

$\frac{\sqrt{3}}{1}$

Step 4

Divide $\sqrt{3}$ by $1$ .

$\sec (θ) = \sqrt{3}$

Step 5

Approximate the result.

$\sec (θ) = \sqrt{3} \approx 1.7320508$