Question

How do you find the exact value of `cos48` using the sum and difference, double angle or half angle formulas?

Answer 1

`cos48^@=1/8(sqrt(30+6sqrt5)+1-sqrt5)`

Explanation:

Let `A=18^@`, hence `5A=90^@` and

`2A=90^@-3A` and `sin2A=sin(90^@-3A)=cos3A`

or `2sinAcosA=4cos^3A-3cosA`

i.e. `cosA(2sinA-4cos^2A+3)=0` and as `cosA=cos18^@!=0`

we have `2sinA-4cos^2A+3=0`

or `2sinA-4(1-sin^2A)+3=0`

or `4sin^2A+2sinA-1=0` and hence

`sinA=(-2+-sqrt(2^2-4*4*(-1)))/8` or `(-2+-2sqrt5)/8` i.e. `(sqrt5-1)/4`

(as `m/_A=18^@`, `sinA` and `cosA` cannot be negative)

and `cosA=sqrt(1-(sqrt5-1)^2/16)=1/4sqrt(16-(5+1-2sqrt5)`

= `sqrt(10+2sqrt5)/4`

And `cos48^@=cos(30^@+18^@)`

= `cos30^@cos18^@-sin30^@sin18^@`

= `sqrt3/2xxsqrt(10+2sqrt5)/4-1/2xx(sqrt5-1)/4`

= `1/8(sqrt(30+6sqrt5)+1-sqrt5)`