Math / Trigonometry

QuestionGiven that $\boldsymbol{\operatorname { t a n }} \phi=2$ , find the other trigonometric functions. $\sin \phi=\frac{2 \sqrt{5}}{5}$ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) $\cos \phi=\square$ $\square$ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Studdy Solution

STEP 1

1. We are given that $\tan \phi = 2$ .
2. We need to find the other trigonometric functions: $\sin \phi$ and $\cos \phi$ .
3. The expression for $\sin \phi$ is provided as $\frac{2 \sqrt{5}}{5}$ .

STEP 2

1. Use the definition of tangent to express $\tan \phi$ in terms of sine and cosine.
2. Use the Pythagorean identity to find $\cos \phi$ .
3. Verify the consistency of the given $\sin \phi$ with the calculated $\cos \phi$ .
4. Simplify the expression for $\cos \phi$ .

STEP 3

Recall the definition of tangent:
$\tan \phi = \frac{\sin \phi}{\cos \phi}$
Given $\tan \phi = 2$ , we have:
$\frac{\sin \phi}{\cos \phi} = 2$

STEP 4

Use the Pythagorean identity:
$\sin^2 \phi + \cos^2 \phi = 1$
Substitute $\sin \phi = \frac{2 \sqrt{5}}{5}$ :
$\left(\frac{2 \sqrt{5}}{5}\right)^2 + \cos^2 \phi = 1$

STEP 5

Calculate $\sin^2 \phi$ :
$\sin^2 \phi = \left(\frac{2 \sqrt{5}}{5}\right)^2 = \frac{4 \times 5}{25} = \frac{20}{25} = \frac{4}{5}$
Substitute back into the Pythagorean identity:
$\frac{4}{5} + \cos^2 \phi = 1$

STEP 6

Solve for $\cos^2 \phi$ :
$\cos^2 \phi = 1 - \frac{4}{5} = \frac{1}{5}$
Take the square root to find $\cos \phi$ :
$\cos \phi = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}$

STEP 7

Verify the consistency of $\sin \phi$ and $\cos \phi$ :
Since $\tan \phi = \frac{\sin \phi}{\cos \phi} = 2$ , we choose the positive value for $\cos \phi$ to maintain the positive ratio:
$\tan \phi = \frac{\frac{2 \sqrt{5}}{5}}{\frac{\sqrt{5}}{5}} = \frac{2 \sqrt{5}}{\sqrt{5}} = 2$
This is consistent with the given $\tan \phi = 2$ .

STEP 8

The value of $\cos \phi$ is:
$\cos \phi = \frac{\sqrt{5}}{5}$
The values of the trigonometric functions are:
$\sin \phi = \frac{2 \sqrt{5}}{5}, \quad \cos \phi = \frac{\sqrt{5}}{5}$

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Studdy Solution

STEP 1

1. We are given that $\tan \phi = 2$ .
2. We need to find the other trigonometric functions: $\sin \phi$ and $\cos \phi$ .
3. The expression for $\sin \phi$ is provided as $\frac{2 \sqrt{5}}{5}$ .

STEP 2

1. Use the definition of tangent to express $\tan \phi$ in terms of sine and cosine.
2. Use the Pythagorean identity to find $\cos \phi$ .
3. Verify the consistency of the given $\sin \phi$ with the calculated $\cos \phi$ .
4. Simplify the expression for $\cos \phi$ .

STEP 3

Recall the definition of tangent:
$\tan \phi = \frac{\sin \phi}{\cos \phi}$
Given $\tan \phi = 2$ , we have:
$\frac{\sin \phi}{\cos \phi} = 2$

STEP 4

Use the Pythagorean identity:
$\sin^2 \phi + \cos^2 \phi = 1$
Substitute $\sin \phi = \frac{2 \sqrt{5}}{5}$ :
$\left(\frac{2 \sqrt{5}}{5}\right)^2 + \cos^2 \phi = 1$

STEP 5

Calculate $\sin^2 \phi$ :
$\sin^2 \phi = \left(\frac{2 \sqrt{5}}{5}\right)^2 = \frac{4 \times 5}{25} = \frac{20}{25} = \frac{4}{5}$
Substitute back into the Pythagorean identity:
$\frac{4}{5} + \cos^2 \phi = 1$

STEP 6

Solve for $\cos^2 \phi$ :
$\cos^2 \phi = 1 - \frac{4}{5} = \frac{1}{5}$
Take the square root to find $\cos \phi$ :
$\cos \phi = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}$

STEP 7

STEP 8

The value of $\cos \phi$ is:
$\cos \phi = \frac{\sqrt{5}}{5}$
The values of the trigonometric functions are:
$\sin \phi = \frac{2 \sqrt{5}}{5}, \quad \cos \phi = \frac{\sqrt{5}}{5}$

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Studdy Solution

STEP 1

1. We are given that tan⁡ϕ=2\tan \phi = 2tanϕ=2.2. We need to find the other trigonometric functions: sin⁡ϕ\sin \phisinϕ and cos⁡ϕ\cos \phicosϕ.3. The expression for sin⁡ϕ\sin \phisinϕ is provided as 255\frac{2 \sqrt{5}}{5}525​​.

STEP 2

STEP 3

Recall the definition of tangent:tan⁡ϕ=sin⁡ϕcos⁡ϕ\tan \phi = \frac{\sin \phi}{\cos \phi}tanϕ=cosϕsinϕ​Given tan⁡ϕ=2\tan \phi = 2tanϕ=2, we have:sin⁡ϕcos⁡ϕ=2\frac{\sin \phi}{\cos \phi} = 2cosϕsinϕ​=2

STEP 4

Use the Pythagorean identity:sin⁡2ϕ+cos⁡2ϕ=1\sin^2 \phi + \cos^2 \phi = 1sin2ϕ+cos2ϕ=1Substitute sin⁡ϕ=255\sin \phi = \frac{2 \sqrt{5}}{5}sinϕ=525​​:(255)2+cos⁡2ϕ=1\left(\frac{2 \sqrt{5}}{5}\right)^2 + \cos^2 \phi = 1(525​​)2+cos2ϕ=1

STEP 5

STEP 6

Solve for cos⁡2ϕ\cos^2 \phicos2ϕ:cos⁡2ϕ=1−45=15\cos^2 \phi = 1 - \frac{4}{5} = \frac{1}{5}cos2ϕ=1−54​=51​Take the square root to find cos⁡ϕ\cos \phicosϕ:cos⁡ϕ=±15=±55\cos \phi = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}cosϕ=±51​​=±55​​

STEP 7

STEP 8

The value of cos⁡ϕ\cos \phicosϕ is:cos⁡ϕ=55\cos \phi = \frac{\sqrt{5}}{5}cosϕ=55​​ The values of the trigonometric functions are:sin⁡ϕ=255,cos⁡ϕ=55\sin \phi = \frac{2 \sqrt{5}}{5}, \quad \cos \phi = \frac{\sqrt{5}}{5}sinϕ=525​​,cosϕ=55​​

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1. We are given that $\tan \phi = 2$ .
2. We need to find the other trigonometric functions: $\sin \phi$ and $\cos \phi$ .
3. The expression for $\sin \phi$ is provided as $\frac{2 \sqrt{5}}{5}$ .

Recall the definition of tangent:
$\tan \phi = \frac{\sin \phi}{\cos \phi}$
Given $\tan \phi = 2$ , we have:
$\frac{\sin \phi}{\cos \phi} = 2$

Use the Pythagorean identity:
$\sin^2 \phi + \cos^2 \phi = 1$
Substitute $\sin \phi = \frac{2 \sqrt{5}}{5}$ :
$\left(\frac{2 \sqrt{5}}{5}\right)^2 + \cos^2 \phi = 1$

Solve for $\cos^2 \phi$ :
$\cos^2 \phi = 1 - \frac{4}{5} = \frac{1}{5}$
Take the square root to find $\cos \phi$ :
$\cos \phi = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}$

The value of $\cos \phi$ is:
$\cos \phi = \frac{\sqrt{5}}{5}$
The values of the trigonometric functions are:
$\sin \phi = \frac{2 \sqrt{5}}{5}, \quad \cos \phi = \frac{\sqrt{5}}{5}$