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.)
This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Sunday, November 11, 2012
Saturday, November 10, 2012
Variable Separation - Arbitrary Constant
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
"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
Friday, November 9, 2012
Simplifying Complex Fraction
Category: Arithmetic
"Published in Newark, California, USA"
Simplify:
Solution:
A complex fraction is a fraction which contains fractions at the numerator and denominator. There's a rule in Mathematics that fractions must be simplified always. A complex fraction must be simplified into a simple fraction. Also, it must be simplified into lowest term. If there are whole numbers and mixed fractions, they must be expressed into improper fractions. For instance, 1 and 2 must be written as
Let's consider the numerator. Their Least Common Denominator (LCD) is 12. The above fraction can be written as
Let's consider the denominator. Their Least Common Denominator (LCD) is 6. The above fraction can be written as
If you divide a fraction with another fraction, the divisor must be written in reciprocal form and perform the multiplication as follows
You noticed that you can cancel 13 and 12 is divisible by 6, the above fraction can be written as
Therefore, the answer is ½
"Published in Newark, California, USA"
Simplify:
Solution:
A complex fraction is a fraction which contains fractions at the numerator and denominator. There's a rule in Mathematics that fractions must be simplified always. A complex fraction must be simplified into a simple fraction. Also, it must be simplified into lowest term. If there are whole numbers and mixed fractions, they must be expressed into improper fractions. For instance, 1 and 2 must be written as
Let's consider the numerator. Their Least Common Denominator (LCD) is 12. The above fraction can be written as
Let's consider the denominator. Their Least Common Denominator (LCD) is 6. The above fraction can be written as
If you divide a fraction with another fraction, the divisor must be written in reciprocal form and perform the multiplication as follows
You noticed that you can cancel 13 and 12 is divisible by 6, the above fraction can be written as
Therefore, the answer is ½
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