Question

How do you use the sum and difference formula to simplify `cos((17pi)/12)`?

Answer 1

`cos((17pi)/12)=(1-sqrt3)/(2sqrt2)`

Explanation:

`cos(A+B)=cosAcosB-sinAsinB`

Let `A=(8pi)/12=2pi/3` and `B=(9pi)/12=(3pi)/4`, then `A+B=(17pi)/12`

Now `cosA=cos((2pi)/3)=cos(pi-pi/3)=-cos(pi/3)=-1/2`;

`sinA=sin((2pi)/3)=sin(pi-pi/3)=sin(pi/3)=sqrt3/2`;

`cosB=cos(3pi)/4=cos(pi-pi/4)=-cospi/4=-1/sqrt2`; and

`sinB=sin((3pi)/4)=sin(pi-pi/4)=sin(pi/4)=1/sqrt2`

Hence, `cos((17pi)/12)=cos(A+B)`

= `cosAcosB-sinAsinB`

= `(-1/2)xx(-1/sqrt2)-(sqrt3/2)xx(1/sqrt2)`

= `1/(2sqrt2)-sqrt3/(2sqrt2)=(1-sqrt3)/(2sqrt2)`

Answer 2

`cos((17pi)/12)`

`cos((24pi-7pi)/12)`

`=cos(2pi-(7pi)/12)`

`=cos((7pi)/12)`

`=cos((4pi+3pi)/12)`

`=cos((4pi)/12+(3pi)/12)`

`=cos(pi/3+pi/4)`

`=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)`

`=1/2*1/sqrt2-sqrt3/2*1/sqrt2`

`=(1-sqrt3)/(2sqrt2)`

`=1/4(sqrt2-sqrt6)`

Answer 3

`- cos ((5pi)/12)`

Explanation:

Simplify -->
`cos ((17pi)/12) = cos ((5pi)/12 + pi) = - cos ((5pi)/12)`

Answer 4

`- sqrt(2 - sqrt3)/2`

Explanation:

`cos((17pi)/12) = cos ((-7pi)/12 + (24pi)/12) = cos ((-7pi)/12 + 2pi) =`
`= cos ((-7pi)/12) = cos ((7pi)/12) = cos (pi/12 + pi/2) = - sin (pi/12)`
Find `sin (pi/12)` by using trig identity:
`2sin^2 (pi/12) = 1 - cos (pi/6) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 (pi/12) = (2 - sqrt3)/4`
`sin (pi/12) = +- sqrt(2 - sqrt3)/2`.
Since `sin (pi/12)` is positive, then take the positive value.
`sin (pi/12) = sqrt(2 - sqrt3)/2`
Finally,
`cos ((17pi)/12) = - sin (pi/12) = - sqrt(2 - sqrt3)/2`