# Trigonometry

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**Problem**

The tide in Bay of Fundy rises and falls every 13 hours. The depth of the water at a certain point in the bay is modeled by a function *d* = 5 sin (2π/13)*t* + 9, where *t* is time in hours and *d* is depth in meters. Find the depth at *t* = 13/4 (high tide) and *t* = 39/4 (low tide).

- The depth of the high tide is 15 meters and the depth of the low tide is 3 meters.
- The depth of the high tide is 16 meters and the depth of the low tide is 2 meters.
- The depth of the high tide is 14 meters and the depth of the low tide is 4 meters.
- The depth of the high tide is 17 meters and the depth of the low tide is 1 meter.

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**Problem**

The number of hours daylight, *D*(*t*) at a particular time of the year can be approximated by

for *t* days and *t* = 0 corresponding to January 1. The constant *K* determines the total variation in day length and depends on the latitude of the locale. When is the day length the longest, assuming that it is NOT a leap year?

A. December 20 | C. June 20 |

B. June 19 | D. December 19 |

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## Derivation of Cosine Law

The following are the formulas for cosine law for any triangles with sides a, b, c and angles A, B, C, respectively.

$b^2 = a^2 + c^2 - 2ac\cos B$

$c^2 = a^2 + b^2 - 2ab\cos C$

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## Derivation of Sine Law

For any triangles with vertex angles and corresponding opposite sides are A, B, C and a, b, c, respectively, the sine law is given by the formula...

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## Derivation of Pythagorean Identities

In reference to the right triangle shown and from the functions of a right triangle:

a/c = sin θ

b/c = cos θ

c/b = sec θ

c/a = csc θ

a/b = tan θ

b/a = cot θ

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## Derivation of Sum and Difference of Two Angles

The sum and difference of two angles can be derived from the figure shown below.

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## Derivation of the Double Angle Formulas

The Double Angle Formulas can be derived from Sum of Two Angles listed below:

$\sin (A + B) = \sin A \, \cos B + \cos A \, \sin B$ → Equation (1)

$\cos (A + B) = \cos A \, \cos B - \sin A \, \sin B$ → Equation (2)

$\tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \, \tan B}$ → Equation (3)

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## Derivation of Pythagorean Theorem

**Pythagorean Theorem**

In any right triangle, the sum of the square of the two perpendicular sides is equal to the square of the longest side. For a right triangle with legs measures $a$ and $b$ and length of hypotenuse $c$, the theorem can be expressed in the form

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