Trigonometry> cosecA + cosec2A + cosec4A + cosec 8A …. ...

cosecA + cosec2A + cosec4A + cosec 8A …. n terms =

Ram Ganesh , 10 Years ago

Grade 10

5 Answers

Arun Kumar

Last Activity: 10 Years ago

If its

$cosecA + cosec2A + cosec4A + cosec 8A.... \\=>cosecA(1+2secA+4secAsec2A+8secAsec2Asec4A......) \\=>cosecA(sinA/sinA+2sinA/sinAcosA+4sinA/sinAcosAcos2A....) \\=>cosecA(sinA/sinA+2^2sinA/sin2A+4^2sinA/sin4A+8^2sinA/sin8A....) \\=>1/sinA+2^2/sin2A+4^2/sin4A+8^2/sin8A....$

Now its doesn't seem to solve into anything.

Arun Kumar

Sanip Panta

Last Activity: 8 Years ago

The answer of this question is: CotA-cotnA °n is the angle of last term as cosecnA. For example:CosecA+cosec2A+cosec4A+cosec8A=cotA-cot8A

Sanip Panta

Last Activity: 8 Years ago

The answer was a bit mistake. Actually it was cotA/2 - cotnA #bA is the angle of last term. For example :CosecA+cosec2A +cosec4A +cosec8A +cosec16A = cotA/2 - cot16A

Shree Ram Gupta

Last Activity: 7 Years ago

= cotA - cot8A= cotA - cos8A/sin8A= cotA - 2cos^2(4A) - 1/2sin4Acos4A= cotA - 2cos^2(4A)/2sin4Acos4A + 1/2sin4Acos4A= cotA - cos4A/sin4A + 1/sin8A= cotA - 2cos^2(2A) - 1/2sin2Acos2A + cosec8A= cotA - 2cos^2(A)/2sin2Acos2A + 1/2sin2Acos2A + cosec8A= cotA - cos2A/sin2A+ 1/sin4A+cosec8AcotA - 2cos^2(A) - 1/2sinAcosA + cosec4A + cosec8A=cosA/sinA - cosA/sinA + 1/sin2A + cosec4A + cosec8A= cosec2A + cosec4A + cosec8A