Category: Trigonometry
"Published in Newark, California, USA"
On November 8, 2012, we derived the formulas for Sine Law and Cosine Law. Right now, we will derive the formula for Tangent Law. Let's consider the formula for Sine Law:
In this derivation, we will use only the first two terms of the given equation above. Let's have this one
Arrange the above equation, we have
Add 1 on both sides of the equation and simplify
Subtract 1 on both sides of the equation and simplify
Divide equation 1 by equation 2
At the right side of the equation, use the Sum and Product of Two Angles Formula to convert the sum of the two angles into the product of two angles, we have
Therefore,
If you will derive using the other term of Sine Law, the other formulas of Tangent Law will be
Category: Trigonometry
"Published in Newark, California, USA"
On October 9, 2012, we derived the formulas for Sum and Difference of Two Angles Formula as follows:
Sin (θ + ϕ) = Sin θ Cos ϕ + Cos θ Sin ϕ (equation 1)
Sin (θ - ϕ) = Sin θ Cos ϕ - Cos θ Sin ϕ (equation 2)
Cos (θ + ϕ) = Cos θ Cos ϕ - Sin θ Sin ϕ (equation 3)
Cos (θ - ϕ) = Cos θ Cos ϕ + Sin θ Sin ϕ (equation 4)
If you add equation 1 and equation 2, the equation becomes
Sin (θ + ϕ) + Sin (θ - ϕ) = 2 Sin θ Cos ϕ
Sin θ Cos ϕ = ½ Sin (θ + ϕ) + ½ Sin (θ - ϕ)
If you subtract equation 1 and equation 2, the equation becomes
Sin (θ + ϕ) - Sin (θ - ϕ) = 2 Cos θ Sin ϕ
Cos θ Sin ϕ = ½ Sin (θ + ϕ) - ½ Sin (θ - ϕ)
If you add equation 3 and equation 4, the equation becomes
Cos (θ + ϕ) + Cos (θ - ϕ) = 2 Cos θ Cos ϕ
Cos θ Cos ϕ = ½ Cos (θ + ϕ) + ½ Cos (θ - ϕ)
If you subtract equation 3 and equation 4, the equation becomes
Cos (θ + ϕ) - Cos (θ - ϕ) = - 2 Sin θ Sin ϕ
Sin θ Sin ϕ = - ½ Cos (θ + ϕ) + ½ Cos (θ - ϕ)
Sin θ Sin ϕ = ½ Cos (θ - ϕ) - ½ Cos (θ + ϕ)
Therefore, the formulas for the transformation of Product to Sum of Two Angles are
Sin θ Cos ϕ = ½ Sin (θ + ϕ) + ½ Sin (θ - ϕ)
Cos θ Sin ϕ = ½ Sin (θ + ϕ) - ½ Sin (θ - ϕ)
Cos θ Cos ϕ = ½ Cos (θ + ϕ) + ½ Cos (θ - ϕ)
Sin θ Sin ϕ = ½ Cos (θ - ϕ) - ½ Cos (θ + ϕ)
(Note: I strongly suggest that you must remember or memorize these formulas because you will use these formulas often in Integral Calculus especially the integration of product of two trigonometric functions with different angles.)
Consider the following formulas:
Sin (θ + ϕ) + Sin (θ - ϕ) = 2 Sin θ Cos ϕ
Sin (θ + ϕ) - Sin (θ - ϕ) = 2 Cos θ Sin ϕ
Cos (θ + ϕ) + Cos (θ - ϕ) = 2 Cos θ Cos ϕ
Cos (θ + ϕ) - Cos (θ - ϕ) = - 2 Sin θ Sin ϕ
If x = θ + ϕ and y = θ - ϕ, the values of θ and ϕ are
x = θ + ϕ x = θ + ϕ
y = θ - ϕ - (y = θ - ϕ)
----------------- -----------------
x + y = 2 θ x - y = 2 ϕ
θ = ½ (x + y) ϕ = ½ (x - y)
Substitute the values of θ + ϕ, θ - ϕ, θ, and ϕ to the four equations above, we have
Sin x + Sin y = 2 Sin ½ (x + y) Cos ½ (x - y)
Sin x - Sin y = 2 Cos ½ (x + y) Sin ½ (x - y)
Cos x + Cos y = 2 Cos ½ (x + y) Cos ½ (x - y)
Cos x - Cos y = - 2 Sin ½ (x + y) Sin ½ (x - y)
Therefore, the formulas for the transformation of Sum to Product of Two Angles are
Sin x + Sin y = 2 Sin ½ (x + y) Cos ½ (x - y)
Sin x - Sin y = 2 Cos ½ (x + y) Sin ½ (x - y)
Cos x + Cos y = 2 Cos ½ (x + y) Cos ½ (x - y)
Cos x - Cos y = - 2 Sin ½ (x + y) Sin ½ (x - y)
(Note: I strongly suggest that you must remember or memorize these formulas because you will use these formulas also for proving of trigonometric identities.)
Category: Differential Equation, Integral Calculus
"Published in Newark, California, USA"
Find the particular solution for the equation:
when x = 0, y = 0.
Solution:
If you examine the given equation, it is a differential equation because of the presence of y'. The above equation can be written as
You notice that the exponent of e has two terms. The above equation can be written as
Next, arrange the above equation by separation of variables
Consider the left side of the equation. If u = -y, then du = -dy.
Consider the right side of the equation. If u = -x2, then du = - 2x dx.
Integrate both sides of the equation, we have
To solve for the value of C, substitute x = 0 and y = 0 to the above equation
Therefore, the particular solution is
