Trigonometry–HardProblems
Solve the problem. This problem is very difficult to understand. Let’s see if we can make sense of it. Note that there are multiple interpretations of the problem and that they are all unsatisfactory.  The problem does not say the lake is a circle, but if it is not, the problem cannot be solved. So, let’s assume the lake is a circle.  What is a transit? From www.surveyhistory.org, we get “The transit is used by the surveyor to measure both horizontal and vertical angles.” A transit A piling
 What is a piling? From wisegeek.com, a piling is “a component in a foundation which is driven into the ground to ensure that the foundation is deep.”  We will need a right angle to solve a problem with only one length and angle given, so let’s infer a 90˚ angle as shown in the following diagram. Based on the diagram, which meets every criterion laid out in the problem, we now have: tan 35°
90
90 ∙ tan 35°
ft. The body of water in this problem would be better described as a pond than a lake! Page 1 of 10
Trigonometry – Hard Problems Based on the illustration at right, we get the following: 90
200
tan
tan
.45
.45 ° The key on this type of problem is to draw the correct triangle. Recall that tan
Since is in 2, we can plot the point 3, 8 to help us. Notice that the hypotenuse must have length: Then, cos
√
√
3
8
√73. The angle is in Q2, but tangent is defined only in Q1 and Q4. Further, tan
So we seek the angle in Q4, where tangent is also 0, with the same tangent value as . Recall that the tangent function has a period of radians. Then, tan
tan
Page 2 of 10
0 in Q2. Trigonometry – Hard Problems 5) Find the exact value of the expression. √
a. cos
b. sin
√
c. tan
√
Know where the primary values for the inverse trig functions are defined. sin
θ is defined in Q1 and Q4. cos
θ is defined in Q1 and Q2. tan
θ is defined in Q1 and Q4. You can think of this as adding two vectors with the bearings identified in the problem. Step 1: Convert each vector into itsi and j components, using reference angles. Let u be a vector of 25 mi. at bearing: N63°W From the diagram at right, θ
90°
63°
25 cos 27°
25 sin 27°
27° 22.2752 11.3498 Let v be a vector of 24 mi. at bearing: S27°W From the diagram at right, φ
90°
27°
63° 24 cos 63°
10.8958 24 sin 63°
21.3842 Step 2: Add the results for the two vectors 〈 22.2752,11.3498 〉 〈 10.8958, 21.3842〉 Step 3: Find the resulting angle and convert it to its “bearing” form (see diagram above right). tan
〈 33.1710, 10.0344〉 Bearing Page 3 of 10
10.0344
33.1710
S 90°
16.8° 16.8° W
. °
Trigonometry – Hard Problems 7) A ship is 50 miles west and 31 miles south of a harbor. What bearing should the Captain set to sail directly to harbor? θ
tan
31.8° φ
90°
31.8°
58.2° . °
For Problems 8 and 9 , use: cos
Harmonic motion equations: with
or
sin
or 0 The spring will start at a ‐value of 8, and oscillate from 8 to 8 over time. A good representation of this would be a cosine curve with lead coefficient 8, because cos 0
1, whereas, sin 0
0. The period of our function is 5 seconds. So, we get: 1
period
1
5
The resulting equation, then, is: 1
2 ∙ 5
2
2
5
Note: both of these functions can be graphed using the Trigonometry app available at www.mathguy.us. Assuming that 0 at time 0, it makes sense to use a sine function for this problem. Since the amplitude is 3 feet, a good representation of this would be a sine curve with lead coefficient 3, because sin 0
0, whereas, cos 0
0. Recalling that 2
, and with we get: The resulting equation, then, is: Page 4 of 10
2 ∙
5 . Trigonometry – Hard Problems cos
1
sin
tan
cos
sin
1
cos
cos
cos
∙
sin
sin
cos
cos
cos
cos
cos
sin
sin
cos
sin
sin
sin
∙ cot
sin
sin
∙
cos
sin
sin
cos
AnswerB
sin 75°
sin
1
1
∙ 150°
2
√3
2
1
cos 150°
2
√
2
Note: the half‐angle formula has a " " sign in front. We use “
because 75° is in Q1, and sine values are positive in Q1. ” for this problem cos
3
8
cos
1 3
∙
2 4
1
cos
2
3
4
Note: the half‐angle formula has a "
√2
2
1
2
" sign in front. We use “
because is in Q1, and cosine values are positive in Q1. Page 5 of 10
√
” for this problem Trigonometry – Hard Problems tan
sin
tan
tan
sin
tan
tan
sin
0 1
0 While tan
0 0, π sin
or sin
1
sin
0 is a solution to the equation 1, tan is undefined at Therefore, the original equation, 1 tan
sin
is undefined at , So tan
. is not a solution to this equation. 2 cos
2 1
2
sin
2
sin
2 sin
2
sin
sin
2 sin
sin 0 sin
2 sin
2
0 When an equation contains more than one function, try to convert it to one that contains only one function. 0 0 1
0 sin
0 or 0, π 2 sin
sin
,
,
,
1
. 0 , Page 6 of 10
Trigonometry – Hard Problems cos
∙
‖ ‖‖ ‖
0°
180° ∙
〈1, 1〉 〈4,5〉 1∙4
a ‖ ‖
1
‖ ‖
4
cos
1
5
√41
√ ∙√
1
cos
√
. ° √82
1 √2 ∙
‖ ‖‖ ‖
1 ∙5
∙
〈 1,2〉 〈3, 3〉 1∙3
b ‖ ‖
‖ ‖
cos
2
1
3
3
∙
‖ ‖‖ ‖
cos
2∙
3
√10
3
9 √5 3√2
√ ∙ √
. ° Page 7 of 10
√
Trigonometry – Hard Problems Substitute cos
3
and 3 3 3 3
0 9
4
9
4
rcos Substitute x
, sin
Substitute cos
4
8
4
8 4
4 4
and 4
8 8 8
0 16
4
16 Page 8 of 10
Trigonometry – Hard Problems a) The process of putting a complex number in polar form is very similar to converting a set of rectangular coordinates to polar coordinates. So, if this process seems familiar, that’s because it is. 3
3√3 3
3√3
3√3
3
tan
6 tan
√3 in 2
2
3
sin
So, the polar form is: 6 cos
b)
5 5√3
5√3
5
5
tan
5√3
10 √3
in 3
3
tan
So, the polar form is: 10 cos
7
6
sin
3 cos15°
sin15°
To take a power of two numbers in polar form, take the power of the ‐value and multiply the angle by the exponent. (This is the essence of DeMoivre’s Theorem.) 3 cos15°
81cis 60°
sin15°
81 cos60°
3 ∙ cis 4 ∙ 15°
sin60°
Page 9 of 10
81
81cis 60° 1
2
√3
2
√
Trigonometry – Hard Problems 3
3 √3 3
3√3
3
tan
Then, 3√3
3
3 √3
216cis2
6 tan
√3 in 2
6cis
216 ∙ cos 2
2
3
6 cis 3 ∙
sin 2
Page 10 of 10