# Use the given values to evaluate (if possible) the remaining trigonometric functions. `sintheta=2/sqrt(5)` and `costheta=1/sqrt(5)`. Verify the...

- Use the given values to evaluate (if possible) the remaining trigonometric functions. `sintheta=2/sqrt(5)` and `costheta=1/sqrt(5)`.
- Verify the identity `(1-2sin^2theta)/(sinthetacostheta)=cottheta-tantheta`.
- Solve the equation on the interval `[0,2pi)`. `sinthetacostheta=-1/2`
- A motorized sailboat leaves Naples, Florida, bound for Key West, 150 miles away. Maintaining a constant speed of 15 miles per hour, but encountering heavy cross-winds and strong currents, the crew finds, after 4 hours, that the sailboat is off course by 20°.

a.) How far is the boat from Key West at this time?

b.) Through what angle `theta` should the sailboat turn to correct its course?

c.) How much time has been added to the trip? (Assuming the boat maintains its speed at 15mph.)

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### 2 Answers

1) Unfortunately, I'm not too sure what you're asking for with this question. However, we can solve for theta easily. Note that there is going to be only a single instance where both sine and cosine are positive (between 0 and `pi/2`) and where the sine and cosine take on those values. To find this value, we simply take the inverse sine or cosine, whichever you prefer, and then ensure it is within the range we know it must be. If it is not within the correct range, then we simply go to the unit circle and reflect the angle vertically or horizontally until it gives us a number within the correct range.

`sintheta=2/sqrt(5)`

`sin^(-1)(sintheta) = 1.11`

Considering our range of 0 to `pi/2=1.57` contains the solution, and our angle is contained within, we can leave **our solution as** `theta=1.11`.

2) Start with the left side and manipulate it with identities as needed to reach the right side, as follows:

`(1-2sintheta)/(sinthetacostheta)`

`(cos2theta)/(sinthetacostheta)`

`(cos^2theta-sin^2theta)/(sinthetacostheta)`

`(cos^2theta)/(sinthetacostheta) - (sin^2theta)/(sinthetacostheta)`

`costheta/sintheta-sintheta/costheta`

`cottheta-tantheta`

Note, the key step there was recognizing the numerator `1-2sin^theta` was an identity for `cos(2theta)`.

3) Start by noticing that this also looks like an identity:

`sinthetacostheta=-1/2`

`1/2 sin2theta = -1/2`

`sin2theta=-1`

So, we are looking for `theta` in `[0,2pi)` that fit this relation. This means that `2theta` will be in the set `[0,4pi)`, or two full revolutions on the unit circle. The commonly noted answer will be the following:

`2theta = 3/2 pi`

`theta = 3/4 pi`

Now, this answer is correct and is within the boundaries. However, considering 2theta has a wider range, we have not considered all solutions. In a second revolution on the unit circle, we get the following additional solution:

`2theta = 2pi+ 3/2 pi = 7/2 pi`

`theta = 7/4 pi`

Therefore, we get two solutions on the interval `[0,2pi)`: `theta in {3/4pi, 7/4pi}`

4) To answer this question, we have to recognize what we are given and how to solve for the rest.

a) If you draw out the situation, you quickly notice we are given two sides of a triangle with the angle in between them, and we are asked for the third side. The law of cosines describes the equation we can use to quickly plug-and-chug our way to the answer:

`d^2=60^2+150^2-2(60)(150)cos(20/180*pi)`

`d^2=26100-18000cos(pi/9)`

`d^2=9185`

`d=95.8`

The distance left to travel will be 95.8 miles.

b) What angle will they have to turn? To do this, we use the law of cosines. It is tempting to use the law of sines, but it may output two possible answers (because the range of `sin^-1theta` is `[-pi/2,pi/2]`). We set up our equation as below:

`150^2=60^2+95.8^2-2(60)(98.5)costheta`

Simplifying:

`-0.84 = costheta`

`theta= 2.57 = 147.1^o`

Therefore, the angle between the two paths will be 147.1 degrees. However, the question asks for the angle to *turn*. Therefore, our turning angle will be 180-147.1 = **32.9 degrees**.

c) How much time has been added to this trip is based on how much time it would have taken without the detour and the expected time with the detour. Based on the speed given, it would have taken 10 hours to travel in a straight line. Our new distance, though is 60 miles along with the 95.8 after the turn, giving us a total distance of 155.5 miles. Therefore, our added distance is 5.5 miles, which would take** 22 extra minutes** at 15 mph.

`1)`

`sin theta= 2/sqrt(5)` `cos theta=1/sqrt(5)`

`sec theta= 1/cos theta=1/(1/sqrt(5))=sqrt(5)` `cosec theta=1/sin theta=1/(2/sqrt(5))=1/2sqrt(5)`

`tan theta= sin theta/cos theta=(cosec theta)/sec theta= (1/2 sqrt(5))/sqrt(5)=1/2`

`cotan theta=1/tan theta= 2`