Sine Law - Cosine Law

Category: Trigonometry

"Published in Newark, California, USA"

Consider the triangle below:

Photo by Math Principles in Everyday Life

Can we get the trigonometric functions of the given triangle? If you will use the trigonometric functions for the right triangle, then it is not applicable for the given triangle because the given triangle is acute triangle and not a right triangle. We can make the given triangle into two right triangles by draw a perpendicular line or altitude from point B to the base which is AC. Let h be the altitude of a triangle as shown in the figure:

Photo by Math Principles in Everyday Life

First, let's derive the Sine Law for the given triangle. Since the given triangle has two right triangles, we can now use the trigonometric functions for the right triangle as follows:

if h = h, then

                                c Sin A = a Sin C

Rearrange the above equation,

If you will do the same procedure by draw a perpendicular line or altitude from points A and C, then the equation is 

Next, let's derive for Cosine Law for the given triangle. The altitude of a triangle divides the base into two parts as shown below:

Photo by Math Principles in Everyday Life

Since the given triangle has two right triangles, we can now use the trigonometric functions for the right triangle as follows:

By Pythagorean Theorem,

                           a² = h² + y²

but                         b = x + y

                              y = b - x

Therefore,

  a² = h² + y²

  a² = (c Sin A)² + (b - x)²

  a² = (c Sin A)² + (b - c Cos A)²

  a² = c² Sin² A + b² - 2bc Cos A + c² Cos² A

  a² = c² (Sin² A + Cos² A) + b² - 2bc Cos A

but       Sin² A + Cos² A = 1

            a² = b² + c² - 2bc Cos A

If you will do the same procedure by draw a perpendicular line or altitude from points A and C, then the equations are

           b² = a² + c² - 2ac Cos B

           c² = a² + b² - 2ab Cos C 

Two Intersecting Lines

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the points of intersection of the following lines:

                          x + 2y - 3 = 0
                         
                          2x - 3y + 8 = 0

Solution:

Since the given equations are all first degree, then they are linear equations. They are straight lines. We can graph the two lines by getting their slope and y-intercept. 

For x + 2y - 3 = 0,

                           x + 2y - 3 = 0
                           2y = -x + 3
                           y = - ½ x + ³/₂

                           slope (∆y/∆x), m = - ½
                           y-intercept = ³/₂

To trace the graph, plot ³/₂ at the y-axis. This is your first point of the line (0, ³/₂). Next, use the slope to get the second point. From the first point, count 2 units to the left and then 1 unit upward. 

For 2x - 3y + 8 = 0,

                          2x - 3y + 8 = 0
                          3y = 2x + 8
                          y = ⅔ x + ⁸/₃

                          slope (∆y/∆x), m = ⅔
                          y-intercept = ⁸/₃ or 2 ⅔

To trace the graph, plot ⁸/₃ at the y-axis. This is your first point of the line (0, ⁸/₃). Next, use the slope to get the second point. From the first point, count three units to the right and then 2 units upward.

Photo by Math Principles in Everyday Life

From the graph, their point of intersection is (-1, 2). To get their actual point of intersection, we have to use the two given equations and solve for x and y, we have

                                 x + 2y - 3 = 0


                                 2x - 3y + 8 = 0

Multiply the 1st equation by 2 and -1 at the 2nd equation. Add the two equations in order to eliminate x and solve for the value of y.

    2(x + 2y - 3 = 0)             2x + 4y - 6 = 0
                                 →
  -1(2x - 3y + 8 = 0)           -2x + 3y - 8 = 0
                                     ________________

                                                7y - 14 = 0
                                                7y = 14
                                                  y = 2

Substitute y to either of the two equations,

                                 x + 2y - 3 = 0
                                 x + 2(2) - 3 = 0
                                 x + 4 - 3 = 0
                                 x + 1 = 0
                                 x = -1

Therefore, their point of intersection is P(-1, 2).

Photo by Math Principles in Everyday Life

Proving Trigonometric Identities - Half Angles

Category: Trigonometry

"Published in Newark, California, USA"

Prove the trigonometric identity for 

Solution:

In proving the trigonometric identities, you need to examine the both sides of the equation very well. You have to choose the more complicated side of the equation and then simplify as much as you can until it is matched with the other side of the equation. In this case, the right side of the equation is more complicated. Let's start with simplifying the right side of the equation as follows

but

and hence

Therefore,