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Alex
3 years ago
9

Write cos(3x) - cos x as a product

Mathematics
1 answer:
sineoko [7]3 years ago
3 0

Answer:

-2sin(x) * sin(2x)

or any equivalent form, such as 4cos(x)[cos²(x)-1]

Step-by-step explanation:

simplify cos(3x) first:

cos(3x) = cos(2x+x) = cos(2x)cos(x) -sin(2x)sin(x)

using trig identities

= [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)

= 2cos³(x) - cos(x) - 2sin²(x)cos(x)

substituting using trig identity sin²(x) + cos²(x) = 1

2cos³(x) - cos(x) - 2[1-cos²(x)]cos(x)

2cos³(x) - cos(x) - 2cos(x)+2cos³(x)

4cos³(x) - 3cos(x)

remember this cos(3x), we still have to subtract cos(x)

4cos³(x) - 3cos(x) - cos(x) = 4cos³(x) - 4cos(x)

we can factor 4cos(x) to write this as a product of:

4cos(x)[cos²(x)-1]

further simplification if you want

trig identity sin²(x) + cos²(x) = 1

simplifying: sin²(x) = 1-cos²(x)

simplifying: -sin²(x) = cos²(x)-1

4cos(x)[cos²(x)-1]

4cos(x)[-sin²(x)]

-4cos(x)sin²(x)

trig identity: sin(2a) = 2cos(a)sin(a)

-2sin(x) * 2cos(x)sin(x)

-2sin(x)*sin(2x)

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Let n=1. Then n^3-n=1^3-1=0. By convention, every non-zero integer n divides 0, so 3\vert n^3-n.

Suppose this relation holds for n=k, i.e. 3\vert k^3-k. We then hope to show it must also hold for n=k+1.

You have

(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)

We assumed that 3\vert k^3-k, and it's clear that 3\vert 3(k^2+k) because 3(k^2+k) is a multiple of 3. This means the remainder upon divides (k+1)^3-(k+1) must be 0, and therefore the relation holds for n=k+1. This proves the statement.
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3 years ago
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Answer:

The table a not represent a proportional relationship between the two quantities

The table b represent a proportional relationship between the two quantities

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

<u><em>Verify each table</em></u>

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k=\frac{D}{C}

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the values of k are equal

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The volume of the rectangular prism can be express in standard polynomial as follows:

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