show that sin 2theta/1-cos 2 theta = cot theta
To prove that sin(2θ) / (1 - cos(2θ)) is equal to cot(θ), we'll use the trigonometric identities.
Let's start by expressing sin(2θ) and cos(2θ) in terms of sin(θ) and cos(θ):
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos²(θ) - sin²(θ)
Substituting these values into the equation sin(2θ) / (1 - cos(2θ)), we have:
2sin(θ)cos(θ) / (1 - (cos²(θ) - sin²(θ)))
To simplify further, we'll expand the denominator:
2sin(θ)cos(θ) / (1 - cos²(θ) + sin²(θ))
Since 1 - cos²(θ) + sin²(θ) is equal to sin²(θ) + (1 - cos²(θ)), we can rewrite the expression as:
2sin(θ)cos(θ) / (sin²(θ) + (1 - cos²(θ)))
Next, we simplify the expression by using the Pythagorean identity sin²(θ) + cos²(θ) = 1:
2sin(θ)cos(θ) / (sin²(θ) + (1 - sin²(θ)))
This simplifies to:
2sin(θ)cos(θ) / 1
Finally, we can rewrite sin(θ)cos(θ) as 1/2 * sin(2θ). Therefore, the expression becomes:
(1/2 * sin(2θ)) / 1
Simplifying further, we have:
(1/2) * sin(2θ)
Recall that the cotangent of θ is equal to cos(θ) / sin(θ). If we substitute θ with 2θ, we get cos(2θ) / sin(2θ), which is equal to cot(2θ).
Therefore, we have shown that sin(2θ) / (1 - cos(2θ)) is equal to cot(θ).
To prove that sin(2θ) / (1 - cos(2θ)) = cot(θ), we will start with the left-hand side (LHS) of the equation and simplify it step by step.
LHS: sin(2θ) / (1 - cos(2θ))
Step 1: Apply the double-angle formulas for sine and cosine.
sin(2θ) = 2sinθcosθ
cos(2θ) = cos^2θ - sin^2θ
Replacing sin(2θ) and cos(2θ) in the LHS equation:
LHS: (2sinθcosθ) / (1 - (cos^2θ - sin^2θ))
Step 2: Simplify the denominator.
1 - (cos^2θ - sin^2θ) = 1 - cos^2θ + sin^2θ
Using the identity: sin^2θ + cos^2θ = 1
1 - cos^2θ + sin^2θ = sin^2θ + (1 - cos^2θ) = sin^2θ + sin^2θ = 2sin^2θ
Now the denominator becomes 2sin^2θ.
LHS: (2sinθcosθ) / 2sin^2θ
Step 3: Cancel out the 2 from the numerator and denominator.
LHS: (sinθcosθ) / sin^2θ
Step 4: Apply the quotient identity for tangent.
sinθcosθ = (sinθ / cosθ)
Also, sin^2θ = (sinθ)^2
LHS: ((sinθ / cosθ) / (sinθ)^2)
Step 5: Simplify the expression further.
((sinθ / cosθ) / (sinθ)^2) = (sinθ / cosθ) * (1 / (sinθ)^2)
Apply the reciprocal identity for sine:
1 / (sinθ)^2 = csc^2θ
LHS: (sinθ / cosθ) * csc^2θ
Step 6: Apply the identity for cotangent.
sinθ / cosθ = cotθ
LHS: cotθ * csc^2θ
Step 7: Apply the reciprocal identity for cotangent.
cotθ = 1 / tanθ
LHS: (1 / tanθ) * csc^2θ
Step 8: Apply the identity for cosecant.
cscθ = 1 / sinθ
LHS: (1 / tanθ) * (1 / sinθ)^2
Step 9: Simplify the expression further.
(1 / tanθ) * (1 / sinθ)^2 = (1 / tanθ) * 1 / (sinθ)^2
Apply the identity for tangent:
1 / tanθ = cotθ
LHS: cotθ * (1 / (sinθ)^2)
Step 10: Finally, simplify the expression.
cotθ * (1 / (sinθ)^2) = cotθ
Thus, we have successfully proved that sin(2θ) / (1 - cos(2θ)) = cot(θ).
The equation you gave is NOT an identity,
Did you mean
sin (2Ø)/(1 - cos^2 Ø) = 2cotØ ?
LS = 2sinØcosØ/(sin^2 Ø)
= 2cosØ/sinØ
= 2cotØ
= RS