Prove that #cosec(x/4)+cosec (x/2)+cosecx=cot(x/8)-cotx# ?

3 Answers
P dilip_k
Apr 28, 2018

#LHS=cosec(x/4)+cosec (x/2)+cosecx#

#=cosec(x/4)+cosec (x/2)+cosecx+cotx-cotx#

#=cosec(x/4)+cosec (x/2)+color(blue)[1/sinx+cosx/sinx]-cotx#

#=cosec(x/4)+cosec (x/2)+color(blue)[(1+cosx)/sinx]-cotx#

#=cosec(x/4)+cosec (x/2)+color(blue)[(2cos^2(x/2))/(2sin(x/2)cos(x/2))]-cotx#

#=cosec(x/4)+cosec (x/2)+color(blue)(cos(x/2)/sin(x/2))-cotx#
#=cosec(x/4)+ color(green)(cosec (x/2)+cot(x/2))-cotx#

#color(magenta)"Proceeding in similar manner as before"#

#=cosec(x/4)+color(green)cot(x/4)-cotx#

#=cot(x/8)-cotx=RHS#

Ratnaker Mehta
Apr 29, 2018

Kindly go through a Proof given in the Explanation.

Explanation:

Setting #x=8y#, we have to prove that,

#cosec2y+cosec4y+cosec8y=coty-cot8y#.

Observe that, #cosec8y+cot8y=1/(sin8y)+(cos8y)/(sin8y)#,

#=(1+cos8y)/(sin8y)#,

#=(2cos^2 4y)/(2sin4ycos4y)#,

#=(cos4y)/(sin4y)#.

#"Thus, "cosec8y+co8y=cot4y [=cot(1/2*8y)]........(star)#.

Adding, #cosec4y#,

#cosec4y+(cosec8y+co8y)=cosec4y+cot4y#,

#=cot(1/2*4y).........[because, (star)]#.

#:. cosec4y+cosec8y+co8y=cot2y#.

Re-adding #cosec2y# and re-using #(star)#,

#cosec2y+(cosec4y+cosec8y+co8y)=cosec2y+cot2y#,

#=cot(1/2*2y)#.

#:.cosec2y+cosec4y+cosec8y+co8y=coty, i.e., #

# cosec2y+cosec4y+cosec8y=coty-cot8y#, as desired!

Abhishek K.
May 1, 2018

Another approach I seem to have learned previously from respected sir dk_ch.

Explanation:

#RHS=cot(x/8)-cotx#

#=cos(x/8)/sin(x/8)-cosx/sinx#

#=(sinx*cos(x/8)-cosx*sin(x/8))/(sinx*sin(x/8))#

#=sin(x-x/8)/(sinx*sin(x/8))=sin((7x)/8)/(sinx*sin(x/8))#

#=(2sin((7x)/8)*cos(x/8))/(2*sin(x/8)*cos(x/8)*sinx)#

#=(sinx+sin((3x)/4))/(sinx*sin(x/4))=cancel(sinx)/(cancel(sinx)*sin(x/4))+(2sin((3x)/4)*cos(x/4))/(sinx*2*sin(x/4)*cos(x/4))#

#=cosec(x/4)+(sinx+sin(x/2))/(sinx*sin(x/2))=cosecx+cosec(x/2)+coesc(x/4)=LHS#

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