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

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dk_ch Share
Oct 29, 2016

#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)#

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Sep 15, 2016

Answer:

#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)#

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