# Functionf:R->R f(x)=cos 3x + cos 5x - cosx f(pi/24)=?

sciencesolve | Certified Educator

Replacing x by pi/24 yields:

f(pi/24) = cos 3pi/24 + cos 5pi/24 - cospi/24

cos 3pi/24 = cos (2pi/24 + pi/24)

Use cos (x+y) = cos x*cos y - sin x * sin y

cos (2pi/24 + pi/24) = cos 2pi/24*cos pi/24 - sin 2pi/24*sin pi/24

cos (2pi/24 + pi/24) = cos pi/12*cos pi/24 - sin pi/12*sin pi/24

cos pi/12 = cos (pi/6)/2 = sqrt((1+cos pi/6)/2)

cos pi/12 = cos (pi/6)/2 = sqrt(2+sqrt3)/2

sin pi/12 = sqrt((1-cos pi/6)/2)

sin pi/12 = sqrt(2-sqrt3)/2

cos pi/24 = cos (pi/12)/2 = sqrt((1+cos pi/12)/2)

cos pi/24 = cos (pi/12)/2 = sqrt(2+sqrt(2+sqrt3))/2

sin pi/24 = sqrt((1-cos pi/12)/2)

sin pi/24 = sqrt(2-sqrt(2+sqrt3))/2

cos (2pi/24 + pi/24) = sqrt(2+sqrt3)*sqrt(2+sqrt(2+sqrt3))/4 - sqrt(2-sqrt3)*sqrt(2-sqrt(2+sqrt3))/4

cos 5pi/24 = cos (4pi/24 + pi/24) = cos 4pi/24*cos pi/24 - sin 4pi/24*sin pi/24

cos 5pi/24 = cos (4pi/24 + pi/24) = cos pi/6*cos pi/24 - sin pi/6*sin pi/24

cos 5pi/24 = sqrt(3(2+sqrt(2+sqrt3))/4 - sqrt(2-sqrt(2+sqrt3))/2

f(pi/24) = sqrt(2+sqrt3)*sqrt(2+sqrt(2+sqrt3))/4 - sqrt(2-sqrt3)*sqrt(2-sqrt(2+sqrt3))/4 + sqrt(3(2+sqrt(2+sqrt3))/4 - sqrt(2-sqrt(2+sqrt3))/2 - sqrt(2+sqrt(2+sqrt3))/2

giorgiana1976 | Student

In order to calculate f(pi/24), first we have to work a bit the analytical expression of the function f(x).

For this purpose, we'll group the first 2 terms and we'll transform the sum between the trigonometric functions into a product (given the fact that we have 2 similar functions, e.g. cos).

cos 3x + cos 5x =2 cos(3x+5x)/2cos(3x-5x)/2=2cos(4x)cos x

Instead of f(x)=cos 3x + cos 5x - cosx, we'll write

f(x)=2cos(4x)cos x - cosx

We have,in this form, the common factor cos x.

f(x)=cos x * [2cos(4x) -1]

So, f(pi/24)=cos (pi/24)*[2cos4*(pi/24) -1]

f(pi/24)=cos (pi/24)*[2 cos(pi/6) -1]=

cos (pi/24)*[2*sqrt3/2-1]

cos(pi/24)=cos[(pi/6)/4]= cos[(pi/12)/2]

cos(pi/12)=cos(pi/6/2)=sqrt[(1+cos(pi/6))/2]=

sqrt[(1 +sqrt(3)/2)/2]= {sqrt[2+sqrt(3)]}/2

cos (pi/24)=cos [(pi/12)/2]=sqrt[1+cos(pi/12)/2]=

=sqrt{2+sqrt[2+sqrt(3)]}/2

f(pi/24)=sqrt{2+sqrt[2+sqrt(3)]}/2 * [sqrt(3) - 1]