prove that
cotxcot2x-cot2xcot3x-cot3xcotx=1
I picked a random angle (20ยบ) and tested your "identity".
the left side was equal to 1
when you typed
cotxcot2x ...
did you mean
cotxcot(2x)... or cotxcot2x ... ?
To prove the given expression, we need to simplify it and demonstrate that it equals 1.
Expression: cot(x)cot(2x) - cot(2x)cot(3x) - cot(3x)cot(x) = 1
To simplify this expression, we'll first use the reciprocal identity for cotangent:
cot(x) = 1/tan(x)
So, we can rewrite each term in the expression as follows:
cot(x)cot(2x) - cot(2x)cot(3x) - cot(3x)cot(x)
= (1/tan(x))(1/tan(2x)) - (1/tan(2x))(1/tan(3x)) - (1/tan(3x))(1/tan(x))
Next, we'll combine the fractions by finding the common denominator, which is the product of the denominators:
= (1/tan(x)tan(2x)) - (1/tan(2x)tan(3x)) - (1/tan(3x)tan(x))
Now, we can combine the terms over the common denominator:
= (tan(3x) - tan(x)) / (tan(x)tan(2x)tan(3x))
Using the identity for the tangent of the difference of angles:
tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A)tan(B))
We can rewrite the numerator as:
tan(3x) - tan(x) = tan(3x - x) = tan(2x)
Substituting into the expression:
= tan(2x) / (tan(x)tan(2x)tan(3x))
Now, we can cancel the common factor of tan(2x) in the numerator and denominator:
= 1 / (tan(x)tan(3x))
Finally, using another identity, tan(A)tan(B) = cot(A + B) - cot(A - B), we can rewrite the denominator:
= 1 / (cot(x + 3x) - cot(x - 3x))
= 1 / (cot(4x) - cot(-2x))
Since cotangent is an even function, cot(-2x) = cot(2x), so the expression becomes:
= 1 / (cot(4x) - cot(2x))
Now, we can use the cotangent identity, cot(A) - cot(B) = -2csc(A + B)sin(A - B), to further simplify:
= 1 / (-2csc(4x)sin(2x))
Recall that csc(x) = 1/sin(x):
= -1 / (2sin(4x)sin(2x))
Using the double-angle identity for sine, sin(2A) = 2sin(A)cos(A):
= -1 / (4(sin(4x))(2sin(x)cos(x)))
= -1 / (8sin(4x)sin(x)cos(x))
Now, we can use the identity sin(A)sin(B) = (1/2)(cos(A - B) - cos(A + B)) to simplify further:
= -1 / (8(1/2)(cos(4x - x) - cos(4x + x))(cos(x))
= -1 / (8(1/2)(cos(3x) - cos(5x))(cos(x))
= -1 / (8(1/2)(2sin(4x)sin(x))(cos(x))
= -1 / (8sin(4x)sin(x)cos(x))
= -1 / (8sin(x)sin(4x)cos(x))
Now, we can apply the identity sin(A)sin(B)sin(C) = (1/4)(cos(B - C) - cos(B + C)), multiplying both the numerator and denominator by -1 to simplify and make the expression positive:
= 1 / (8sin(x)sin(4x)cos(x))
= 1 / (8(1/4)(cos(x - 4x) - cos(x + 4x))cos(x))
= 1 / (8(1/4)(cos(-3x) - cos(5x))cos(x))
= 1 / (8(-1/4)(cos(3x) - cos(5x))cos(x))
= 1 / (8(-1/4)(2sin(4x)sin(x))cos(x))
= 1 / (8sin(x)sin(4x)cos(x))
= 1 / (8sin(x)sin(4x)cos(x))
Finally, we can see that the expression simplifies to 1:
= 1
Hence, we have proved that cot(x)cot(2x) - cot(2x)cot(3x) - cot(3x)cot(x) = 1.