Find the sine, cosine, and tangent of each angle using the unit circle. θ = − 1.18 \theta=-1.18 θ = − 1.18 rad, ( 5 13 , − 12 13 ) \left(\frac{5}{13},-\frac{12}{13}\right) ( 13 5 , − 13 12 )

In this problem, we're asked to find the sine, cosine, and tangent of each angle using the unit circle. And here we have our given angle as negative one point 18 radians. So our value of theta, our angle is negative 1.18 radians. Now remember whenever we're working with a negative angle, that just means that instead of going counterclockwise around my circle, I'm instead just going clockwise, so nothing to worry about here. Now let's go ahead and look at our unit circle and determine what exactly these trig values are. Now looking here, I have the point 5 over 13 and negative 12 over 13. Now remember, on the unit circle, this ordered pair will tell me my values of cosine and sine. So looking at this, 5 over 13 and negative 12 over 13, remember I have my x y values which correspond to my cosine and sine values, thinking about that in alphabetical order. So here I have this 5 over 13 that I can fill in for my cosine. So the cosine of negative 1.18 radians is 5 over 13. Now my sine is my y value so don't forget that negative here. So my sine of negative one point 18 radians is going to be negative 12 over 13. Now a little bit more work with our tangent here, but nothing too crazy because remember the tangent of our angle is going to be equal to the sine divided by the cosine. Or we can also think about this in terms of our ordered pairs, y over x. So here I can take my y value, which is negative 12 over 13 and divide it by my x value which is 5 over 13. Now this looks a little crazy a fraction divided by a fraction and you can think about this 2 different ways. So the first way you can think of is, okay, if I'm dividing by a fraction, I'm really just multiplying by the reciprocal of the denominator. But here, since these have the same denominators, we can also think of this as just dividing those numerators because those thirteens will end up canceling out. So this fraction is really negative 12 over 5. So the tangent of negative 1.18 radians is going to be a negative 12 over 5. Make sure you don't forget that negative. So that's all for this one. Thanks for watching, and I'll see you in the next one.

Find the sine, cosine, and tangent of each angle using the unit circle.θ=−1.18\theta=-1.18θ=−1.18 rad, (513,−1213)\left(\frac{5}{13},-\frac{12}{13}\right)(135,−1312)

sin⁡θ=−1213,cos⁡θ=513,tan⁡θ=125\sin\theta=-\frac{12}{13},\cos\theta=\frac{5}{13},\tan\theta=\frac{12}{5}

3. Unit Circle

Trigonometric Functions on the Unit Circle

Trigonometry

3. Unit Circle

Trigonometric Functions on the Unit Circle

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Multiple Choice

Find the sine, cosine, and tangent of each angle using the unit circle.
$$\theta$=-1.18$ rad, $$\left$($\frac{5}{13}$,-$\frac{12}{13}\]\right$)$

$\sin θ = - \frac{12}{13}, \cos θ = \frac{5}{13}, \tan θ = \frac{12}{5}$

$\sin\]\theta\)=-$\frac{12}{13}$,$\cos\[\theta$=$\frac{5}{13}$,$\tan\]\theta$=-\(\frac{12}{5}$

$\sin\]\theta\)=$\frac{12}{13}$,$\cos\[\theta$=$\frac{5}{13}$,$\tan\]\theta$=\(\frac{5}{12}$

$\sin\]\theta\)=$\frac{5}{13}$,$\cos\[\theta$=$\frac{-12}{13}$,$\tan\]\theta$=\(\frac{5}{12}$

Verified step by step guidance

Identify the coordinates of the point on the unit circle corresponding to the angle θ = -1.18 rad. From the image, the coordinates are ($\frac{5}{13}$, -$\frac{12}{13}$).

Recall that on the unit circle, the x-coordinate of a point is the cosine of the angle, and the y-coordinate is the sine of the angle. Therefore, $\cos$($\theta$) = $\frac{5}{13}$ and $\sin$($\theta$) = -$\frac{12}{13}$.

To find the tangent of the angle, use the identity $\tan$($\theta$) = $\frac{\sin(\theta)}{\cos(\theta)}$. Substitute the values: $\tan$($\theta$) = $\frac{-\frac{12}{13}$}{$\frac{5}{13}$}.

Simplify the expression for tangent: $\tan$($\theta$) = $\frac{-12}{13}$ $\times$ $\frac{13}{5}$ = -$\frac{12}{5}$.

Verify the results: $\sin$($\theta$) = -$\frac{12}{13}$, $\cos$($\theta$) = $\frac{5}{13}$, $\tan$($\theta$) = -$\frac{12}{5}$. These match the correct answer provided in the problem statement.

6:34m

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(Modeling) Length of a Sag Curve When a highway goes downhill and then uphill, it has a sag curve. Sag curves are designed so that at night, headlights shine sufficiently far down the road to allow a safe stopping distance. See the figure. S and L are in feet. The minimum length L of a sag curve is determined by the height h of the car's headlights above the pavement, the downhill grade θ₁ < 0°, the uphill grade θ₂ > 0°, and the safe stopping distance S for a given speed limit. In addition, L is dependent on the vertical alignment of the headlights. Headlights are usually pointed upward at a slight angle α above the horizontal of the car. Using these quantities, for a 55 mph speed limit, L can be modeled by the formula (θ₂ - θ₁)S² L = ————————— , 200(h + S tan α) where S < L. (Data from Mannering, F., and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.) Compute length L, to the nearest foot, if h = 1.9 ft, α = 0.9°, θ₁ = -3°, θ₂ = 4°, and S = 336 ft.

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Multiple Choice

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Find the sine, cosine, and tangent of each angle using the unit circle.θ=−1.18\(\theta\)=-1.18θ=−1.18 rad, (513,−1213)\(\left\)(\(\frac{5}{13}\),-\(\frac{12}{13}\]\right\))(135,−1312)

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Trigonometric Functions on the Unit Circle practice set

Find the sine, cosine, and tangent of each angle using the unit circle.
$\(\theta\)=-1.18$ rad, $\(\left\)(\(\frac{5}{13}\),-\(\frac{12}{13}\]\right\))$