Prove that
cos (a+b) + sin (a-b) = 2sin(45+a).cos(45+b)

Step-by-step explanation:

cos (A+B) + sin (A-B) = 2 sin (45°+A) cos (45° + B)

= 2 (sin45°cosA + cos45°sinA)(cos45°cosB - sin45°sinB)

But sin45=cos45 =(sqrt2)/2

= 2 ((sqrt2)/2 *cosA + (sqrt2)/2 *sinA)((sqrt2)/2 *cosB -(sqrt2)/2 *sinB)

= 2 ((sqrt2)/2 *(cosA + sinA))*((sqrt2)/2 *(cosB - sinB))

= 2*(sqrt2)/2 * (sqrt2)/2 * (cosA + sinA)*(cosB - sinB)

= (cosA + sinA)*(cosB - sinB)

= cosAcosB +sinAcosB -cosAsinB - sinAsinB

Regrouping:

= (cosAcosB- sinAsinB) + (sinAcosB -cosAsinB)

= cos (A+B) + sin (A-B)

19

step-by-step explanation:

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!