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Solved on Jan 29, 2024

Find the trigonometric ratios and angle θ\theta when cos⁡θ=−420\cos \theta = \frac{-4}{\sqrt{20}} and 0∘≤θ≤180∘0^{\circ} \leq \theta \leq 180^{\circ}.

STEP 1

Assumptions
1. We are given that cos⁡θ=−420\cos \theta=\frac{-4}{\sqrt{20}}.
2. The angle θ\theta is in the range 0∘≤θ≤180∘0^{\circ} \leq \theta \leq 180^{\circ}, which places it in the first or second quadrant.
3. We need to find the six trigonometric ratios: sin⁡θ\sin \theta, cos⁡θ\cos \theta, tan⁡θ\tan \theta, csc⁡θ\csc \theta, sec⁡θ\sec \theta, and cot⁡θ\cot \theta.
4. We will use the Pythagorean identity to find sin⁡θ\sin \theta and then use the definitions of the other trigonometric functions to find the remaining ratios.
5. To find the measure of θ\theta, we will use the inverse trigonometric function of cosine.

STEP 2

Since cos⁡θ=−420\cos \theta=\frac{-4}{\sqrt{20}}, we can simplify the denominator by multiplying the numerator and the denominator by 20\sqrt{20} to get rid of the square root in the denominator.
cos⁡θ=−420×2020\cos \theta=\frac{-4}{\sqrt{20}} \times \frac{\sqrt{20}}{\sqrt{20}}

STEP 3

Simplify the expression by carrying out the multiplication.
cos⁡θ=−42020\cos \theta=\frac{-4\sqrt{20}}{20}

STEP 4

Reduce the fraction by dividing both the numerator and the denominator by 4.
cos⁡θ=−205\cos \theta=\frac{-\sqrt{20}}{5}

STEP 5

Now, we will use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find sin⁡θ\sin \theta.
sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta

STEP 6

Substitute the value of cos⁡θ\cos \theta into the equation.
sin⁡2θ=1−(−205)2\sin^2 \theta = 1 - \left(\frac{-\sqrt{20}}{5}\right)^2

STEP 7

Simplify the right side of the equation.
sin⁡2θ=1−2025\sin^2 \theta = 1 - \frac{20}{25}

STEP 8

Further simplify the right side of the equation.
sin⁡2θ=2525−2025\sin^2 \theta = \frac{25}{25} - \frac{20}{25}

STEP 9

Subtract the fractions.
sin⁡2θ=525\sin^2 \theta = \frac{5}{25}

STEP 10

Reduce the fraction to its simplest form.
sin⁡2θ=15\sin^2 \theta = \frac{1}{5}

STEP 11

Take the square root of both sides to find sin⁡θ\sin \theta. Since 0∘≤θ≤180∘0^{\circ} \leq \theta \leq 180^{\circ}, sin⁡θ\sin \theta will be positive in both the first and second quadrants.
sin⁡θ=15\sin \theta = \sqrt{\frac{1}{5}}

STEP 12

Simplify the square root.
sin⁡θ=55\sin \theta = \frac{\sqrt{5}}{5}

STEP 13

Now that we have sin⁡θ\sin \theta and cos⁡θ\cos \theta, we can find tan⁡θ\tan \theta using the definition tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

STEP 14

Substitute the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta into the equation.
tan⁡θ=55−205\tan \theta = \frac{\frac{\sqrt{5}}{5}}{\frac{-\sqrt{20}}{5}}

STEP 15

Simplify the expression by multiplying the numerator and the denominator by 5.
tan⁡θ=5−20\tan \theta = \frac{\sqrt{5}}{-\sqrt{20}}

STEP 16

Simplify the expression by multiplying the numerator and the denominator by 20\sqrt{20}.
tan⁡θ=5⋅20−20⋅20\tan \theta = \frac{\sqrt{5} \cdot \sqrt{20}}{-\sqrt{20} \cdot \sqrt{20}}

STEP 17

Simplify the expression.
tan⁡θ=100−20\tan \theta = \frac{\sqrt{100}}{-20}

STEP 18

Simplify the square root.
tan⁡θ=10−20\tan \theta = \frac{10}{-20}

STEP 19

Reduce the fraction to its simplest form.
tan⁡θ=−12\tan \theta = -\frac{1}{2}

STEP 20

To find csc⁡θ\csc \theta, use the definition csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}.
csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}

STEP 21

Substitute the value of sin⁡θ\sin \theta.
csc⁡θ=155\csc \theta = \frac{1}{\frac{\sqrt{5}}{5}}

STEP 22

Simplify the expression by multiplying the numerator and the denominator by 5.
csc⁡θ=55\csc \theta = \frac{5}{\sqrt{5}}

STEP 23

Multiply the numerator and the denominator by 5\sqrt{5} to rationalize the denominator.
csc⁡θ=555\csc \theta = \frac{5\sqrt{5}}{5}

STEP 24

Simplify the expression.
csc⁡θ=5\csc \theta = \sqrt{5}

STEP 25

To find sec⁡θ\sec \theta, use the definition sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}.
sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}

STEP 26

Substitute the value of cos⁡θ\cos \theta.
sec⁡θ=1−205\sec \theta = \frac{1}{\frac{-\sqrt{20}}{5}}

STEP 27

Simplify the expression by multiplying the numerator and the denominator by 5.
sec⁡θ=5−20\sec \theta = \frac{5}{-\sqrt{20}}

STEP 28

Multiply the numerator and the denominator by 20\sqrt{20} to rationalize the denominator.
sec⁡θ=520−20\sec \theta = \frac{5\sqrt{20}}{-20}

STEP 29

Simplify the expression.
sec⁡θ=−204\sec \theta = \frac{-\sqrt{20}}{4}

STEP 30

To find cot⁡θ\cot \theta, use the definition cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}.
cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}

STEP 31

Substitute the value of tan⁡θ\tan \theta.
cot⁡θ=1−12\cot \theta = \frac{1}{-\frac{1}{2}}

STEP 32

Simplify the expression.
cot⁡θ=−2\cot \theta = -2
The six trigonometric ratios are:
sin⁡θ=55\sin \theta = \frac{\sqrt{5}}{5} cos⁡θ=−205\cos \theta = \frac{-\sqrt{20}}{5} tan⁡θ=−12\tan \theta = -\frac{1}{2} csc⁡θ=5\csc \theta = \sqrt{5} sec⁡θ=−204\sec \theta = \frac{-\sqrt{20}}{4} cot⁡θ=−2\cot \theta = -2

STEP 33

To determine the measure of θ\theta, we will use the inverse cosine function, since we have the value of cos⁡θ\cos \theta.
θ=cos⁡−1(−205)\theta = \cos^{-1}\left(\frac{-\sqrt{20}}{5}\right)

STEP 34

Use a calculator to find the value of θ\theta to the nearest degree.
θ≈cos⁡−1(−25)\theta \approx \cos^{-1}\left(-\frac{2}{\sqrt{5}}\right)

STEP 35

The calculator will give the value of θ\theta in degrees.
θ≈104∘\theta \approx 104^{\circ}
To the nearest degree, the measure of θ\theta is 104∘104^{\circ}.

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