prove that: (cos(2X) + cosX)/ (sin (2X) - sinX) = cotX+cosecX?

Answer 1

(cos(2x) + cosx)/(sin(2x) - sin(x) = cot(x) + csc(x)

Let's start with the left hand side.

LHS = (cos(2x) + cos(x))/(sin(2x) - sin(x))
= {2cos²(x) - 1 + cos(x)}/{2sin(x)cos(x) - sin(x)}
= (2cos(x) - 1)(cos(x) + 1) / {sin(x)[2cos(x) - 1]}
{cos(x) + 1) / sin(x) = cot(x) + csc(x) = Right Hand Side

Answer 2

If you know how to program in Matlab, for example... or C++, or Java... then you can easily implement this equation and see if the results are the same...

Good luck...
I recommend to you Matlab, if you never saw how a program in Java or C++ looks like. If you want, I can check this thing tomorrow for you.

BUT:
What is cot(x)...? Is it "ctg" (co-tangent)...?
What is cosec(x)? I don't recognize it... Abbreviations slightly differ from country to country...

Answer 3

(cos(2x) + cos(x))/(sin(2x) - sin(x))
(cos(2x) + cos(x))/(2sin(x)cos(x) - sin(x))
(2cos(x)^2 - 1 + cos(x))/((sin(x))(2cos(x) - 1)))
(2cos(x)^2 + cos(x) - 1)/((sin(x))(2cos(x) - 1)))
((cos(x) + 1)(2cos(x) - 1))/((sin(x))(2cos(x) - 1)))
(cos(x) + 1)/(sin(x))

This can also be written as

(cos(x)/sin(x)) + (1/(sin(x)))

which can also be written as

cot(x) + csc(x)

so

(cos(2x) + cos(x))/(sin(2x) - sin(x)) = cot(x) + csc(x)

by the way, you may already know this cosec(x) is the same as writting csc(x)

Answer 4

NO.