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

3 Answers
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

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!

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