Trigonometry Examples

$\sin (θ) = \frac{1}{4}$ , $\tan (θ) > 0$

Step 1

The tangent function is positive in the first and third quadrants. The sine function is positive in the first and second quadrants. The set of solutions for $θ$ are limited to the first quadrant since that is the only quadrant found in both sets.

Solution is in the first quadrant.

Step 2

Use the definition of sine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.

$\sin (θ) = \frac{opposite}{hypotenuse}$

Step 3

Find the adjacent side of the unit circle triangle. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side.

$Adjacent = \sqrt{{hypotenuse}^{2} - {opposite}^{2}}$

Step 4

Replace the known values in the equation.

$Adjacent = \sqrt{{(4)}^{2} - {(1)}^{2}}$

Step 5

Simplify inside the radical.

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Step 5.1

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

Adjacent $= \sqrt{16 - {(1)}^{2}}$

Step 5.2

One to any power is one.

Adjacent $= \sqrt{16 - 1 \cdot 1}$

Step 5.3

Multiply $- 1$ by $1$ .

Adjacent $= \sqrt{16 - 1}$

Step 5.4

Subtract $1$ from $16$ .

Adjacent $= \sqrt{15}$

Step 6

Find the value of cosine.

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Step 6.1

Use the definition of cosine to find the value of $\cos (θ)$ .

$\cos (θ) = \frac{adj}{hyp}$

Step 6.2

Substitute in the known values.

$\cos (θ) = \frac{\sqrt{15}}{4}$

Step 7

Find the value of tangent.

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Step 7.1

Use the definition of tangent to find the value of $\tan (θ)$ .

$\tan (θ) = \frac{opp}{adj}$

Step 7.2

Substitute in the known values.

$\tan (θ) = \frac{1}{\sqrt{15}}$

Step 7.3

Simplify the value of $\tan (θ)$ .

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Step 7.3.1

Multiply $\frac{1}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\tan (θ) = \frac{1}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}}$

Step 7.3.2

Combine and simplify the denominator.

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Step 7.3.2.1

Multiply $\frac{1}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\tan (θ) = \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 7.3.2.2

Raise $\sqrt{15}$ to the power of $1$ .

$\tan (θ) = \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 7.3.2.3

Raise $\sqrt{15}$ to the power of $1$ .

$\tan (θ) = \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 7.3.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\tan (θ) = \frac{\sqrt{15}}{{\sqrt{15}}^{1 + 1}}$

Step 7.3.2.5

Add $1$ and $1$ .

$\tan (θ) = \frac{\sqrt{15}}{{\sqrt{15}}^{2}}$

Step 7.3.2.6

Rewrite ${\sqrt{15}}^{2}$ as $15$ .

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Step 7.3.2.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{15}$ as $15^{\frac{1}{2}}$ .

$\tan (θ) = \frac{\sqrt{15}}{{(15^{\frac{1}{2}})}^{2}}$

Step 7.3.2.6.2

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

$\tan (θ) = \frac{\sqrt{15}}{15^{\frac{1}{2} \cdot 2}}$

Step 7.3.2.6.3

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

$\tan (θ) = \frac{\sqrt{15}}{15^{\frac{2}{2}}}$

Step 7.3.2.6.4

Cancel the common factor of $2$ .

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Step 7.3.2.6.4.1

Cancel the common factor.

$\tan (θ) = \frac{\sqrt{15}}{15^{\frac{2}{2}}}$

Step 7.3.2.6.4.2

Rewrite the expression.

$\tan (θ) = \frac{\sqrt{15}}{15}$

Step 7.3.2.6.5

Evaluate the exponent.

$\tan (θ) = \frac{\sqrt{15}}{15}$

Step 8

Find the value of cotangent.

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Step 8.1

Use the definition of cotangent to find the value of $\cot (θ)$ .

$\cot (θ) = \frac{adj}{opp}$

Step 8.2

Substitute in the known values.

$\cot (θ) = \frac{\sqrt{15}}{1}$

Step 8.3

Divide $\sqrt{15}$ by $1$ .

$\cot (θ) = \sqrt{15}$

Step 9

Find the value of secant.

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Step 9.1

Use the definition of secant to find the value of $\sec (θ)$ .

$\sec (θ) = \frac{hyp}{adj}$

Step 9.2

Substitute in the known values.

$\sec (θ) = \frac{4}{\sqrt{15}}$

Step 9.3

Simplify the value of $\sec (θ)$ .

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Step 9.3.1

Multiply $\frac{4}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\sec (θ) = \frac{4}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}}$

Step 9.3.2

Combine and simplify the denominator.

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Step 9.3.2.1

Multiply $\frac{4}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\sec (θ) = \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 9.3.2.2

Raise $\sqrt{15}$ to the power of $1$ .

$\sec (θ) = \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 9.3.2.3

Raise $\sqrt{15}$ to the power of $1$ .

$\sec (θ) = \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Step 9.3.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sec (θ) = \frac{4 \sqrt{15}}{{\sqrt{15}}^{1 + 1}}$

Step 9.3.2.5

Add $1$ and $1$ .

$\sec (θ) = \frac{4 \sqrt{15}}{{\sqrt{15}}^{2}}$

Step 9.3.2.6

Rewrite ${\sqrt{15}}^{2}$ as $15$ .

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Step 9.3.2.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{15}$ as $15^{\frac{1}{2}}$ .

$\sec (θ) = \frac{4 \sqrt{15}}{{(15^{\frac{1}{2}})}^{2}}$

Step 9.3.2.6.2

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

$\sec (θ) = \frac{4 \sqrt{15}}{15^{\frac{1}{2} \cdot 2}}$

Step 9.3.2.6.3

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

$\sec (θ) = \frac{4 \sqrt{15}}{15^{\frac{2}{2}}}$

Step 9.3.2.6.4

Cancel the common factor of $2$ .

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Step 9.3.2.6.4.1

Cancel the common factor.

$\sec (θ) = \frac{4 \sqrt{15}}{15^{\frac{2}{2}}}$

Step 9.3.2.6.4.2

Rewrite the expression.

$\sec (θ) = \frac{4 \sqrt{15}}{15}$

Step 9.3.2.6.5

Evaluate the exponent.

$\sec (θ) = \frac{4 \sqrt{15}}{15}$

Step 10

Find the value of cosecant.

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Step 10.1

Use the definition of cosecant to find the value of $\csc (θ)$ .

$\csc (θ) = \frac{hyp}{opp}$

Step 10.2

Substitute in the known values.

$\csc (θ) = \frac{4}{1}$

Step 10.3

Divide $4$ by $1$ .

$\csc (θ) = 4$

Step 11

This is the solution to each trig value.

$\sin (θ) = \frac{1}{4}$

$\cos (θ) = \frac{\sqrt{15}}{4}$

$\tan (θ) = \frac{\sqrt{15}}{15}$

$\cot (θ) = \sqrt{15}$

$\sec (θ) = \frac{4 \sqrt{15}}{15}$

$\csc (θ) = 4$