Rewrite the product as a sum: 10cos(5x)sin(10x)

10cos(5x)sin(10x) =  5[sin (15x) + sin (5x)]

Step-by-step explanation:

In this question, we are tasked with writing the product as a sum.

To do this, we shall be using the sum to product formula below;

cosαsinβ = 1/2[ sin(α + β) - sin(α - β)]

From the question, we can say α= 5x and β= 10x

Plugging these values into the equation, we have

10cos(5x)sin(10x) = (10) × 1/2[sin (5x + 10x) - sin(5x - 10x)]

= 5[sin (15x) - sin (-5x)]

We apply odd identity i.e sin(-x) = -sinx

Thus applying same to sin(-5x)

sin(-5x) = -sin(5x)

Thus;

5[sin (15x) - sin (-5x)] = 5[sin (15x) -(-sin(5x))]

= 5[sin (15x) + sin (5x)]

Hence,  10cos(5x)sin(10x) =  5[sin (15x) + sin (5x)]

rounding to the nearest thousandth would be 452,000

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!