Question

How do you use the sum and difference formula to simplify `cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)`?

Answer 1

`cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)=(3+sqrt3)/4`

Explanation:

`cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)`

= `cos(pi/12)[cos((5pi)/12)+sin((5pi)/12)]`

Now as `sin(pi/4)=cos(pi/4)=1/sqrt2` above can be written as

`sqrt2cos(pi/12)[sin(pi/4)cos((5pi)/12)+cos(pi/4)sin((5pi)/12)]`

using `sin(A+B)=sinAcosB+cosAsinB`, this becomes

`sqrt2cos(pi/12)[sin{(pi/4)+((5pi)/12)}`

= `sqrt2cos(pi/12)sin((8pi)/12)=sqrt2sin((8pi)/12)cos(pi/12)`

Now using `sinAcosB=1/2{sin(A-B)+sin(A+B)}`

`sqrt2sin((8pi)/12)cos(pi/12)=sqrt2/2(sin((8pi)/12-pi/12)+sin((8pi)/12+pi/12))`

= `1/sqrt2(sin((7pi)/12)+sin((9pi)/12))`

= `1/sqrt2(sin((3pi)/12+(4pi)/12)+sin((3pi)/4))`

= `1/sqrt2(sin((3pi)/12+(4pi)/12)+1/sqrt2)`

= `1/sqrt2(sin((3pi)/12)cos((4pi)/12)+cos((3pi)/12)sin((4pi)/12))+1/2`

= `1/sqrt2(sin(pi/4)cos(pi/3)+cos(pi/4)sin(pi/3))+1/2`

= `1/sqrt2(1/sqrt2xx1/2+1/sqrt2xxsqrt3/2)+1/2`

= `(sqrt3+1)/4+1/2`

= `(3+sqrt3)/4`