Proving Trigonometric Identities, 7

Category: Trigonometry, Algebra

"Published in Suisun City, California, USA"

Prove that 

Solution:

In proving the trigonometric functions, the first thing that you have to do is to choose the more complicated side of the equation and then simplify and compare with the other side of the equation if they are equal or not. 

In this case, let's choose the left side of the equation as follows

If you will continue to expand further the above equation, it will be more complicated and longer. Let's substitute all  trigonometric functions with another variables as follows

Let 

 

then the above equation becomes

since

then the above equation becomes

Therefore,

Proving Trigonometric Identities, 6

Category: Trigonometry

"Published in Suisun City, California, USA"

Prove that

Solution:

Consider the given equation above

In proving the trigonometric identities, we have to choose the most complicated part which is the left side of the given equation. Let's simplify the left side of the equation. Get the Least Common Denominator (LCD) of the two fractions and rewrite the fractions, we have

   

but 

and the above equation becomes

Therefore,

Permutation Problems, 2

Category: Algebra, Statistics

"Published in Suisun City, California, USA"

A man bought three vanilla ice cream cones, two chocolate cones, four strawberry cones, and five butterscotch cones for his 14 children. In how many ways can he distribute  the cones among his children?

Solution:

The given word problem above is about permutations but it is a different type which is called a Distinguishable Permutation.  

If a set of n objects consists of k different kinds of objects with n₁, objects of the first kind, n₂ objects of the second kind, n₃ objects of the third kind, and so on, where n₁ + n₂ + ......... + n_k = n, then the number of distinguishable permutations of these objects is 

Now, let's go back to the given problem, if n = 14 children, n₁ = 3 vanilla ice cream cones, n₂ = 2 chocolate cones, n3 = 4 strawberry cones, and n₄ = 5 butterscotch cones, then the number of ways to distribute the cones among to his children is