Example 2 :

Example 2: A plane flies at 300 miles per hour. There is a wind blowing out of the southeast at 86 miles per hour with a bearing of 320°. At what bearing must the plane head in order to have a true bearing (relative to the ground) of 14°? What will be the plane’s groundspeed (Figure… Continue reading Example 2 :

Vector Operations

In the physical world, some quantities, such as mass, length, age, and value, can be represented by only magnitude. Other quantities, such as speed and force, also involve direction. You can use vectors to represent those quantities that involve both magnitude and direction. One common use of vectors involves finding the actual speed and direction… Continue reading Vector Operations

Examples

Example 4: Find a unit vector v with the same direction as the vector u given that u = ⟨7, − 1⟩. Two special unit vectors, i = ⟨1, 0⟩ and j = ⟨0, 1⟩, can be used to express any vector v = ⟨a, b⟩. Example 5: Write u = ⟨5, 3⟩ in terms of the i and j unit vectors (Figure 5 ). Figure 5              Drawing for Example 5.… Continue reading Examples

Coordinate System 2

Figure 2               Drawing for Example 1. If the coordinates of point P are ( x, y), An algebraic vector is an ordered pair of real numbers. An algebraic vector that corresponds to standard geometric vector  is denoted as ⟨ a, b⟩ if terminal point P has coordinates of (a, b). The numbers a and b are called the components of vector ⟨ a, b⟩ (see Figure… Continue reading Coordinate System 2

Rectangular Coordinate System

The following discussion is limited to vectors in a two‐dimensional coordinate plane, although the concepts can be extended to higher dimensions. If vector  is shifted so that its initial point is at the origin of the rectangular coordinate plane, it is said to be in standard position. If vector  is equal to vector  and has its initial point at… Continue reading Rectangular Coordinate System

Inverse Trigonometry Formulas

sin-1 (–x) = – sin-1 x cos-1 (–x) = π – cos-1 x tan-1 (–x) = – tan-1 x cosec-1 (–x) = – cosec-1 x sec-1 (–x) = π – sec-1 x cot-1 (–x) = π – cot-1 x

Half Angle Formula of Cos Derivation

Now, we will prove the half angle formula for the cosine function. Using one of the above formulas of cos A, cos A = 2 cos2(A/2) – 1 From this, 2 cos2(A/2) = 1 + cos A cos2 (A/2) = (1 + cos A) / 2 cos (A/2) = ±√[(1 + cos A) / 2]