prove 2csc(2x)= csc(x)^2tanx
L.H. Side:
2 csc(2x) = 2/sin(2x)
= 1/(sinx cosx)
R.H. side:
csc(x)^2 tan x = (1/sin^2x)(sinx/cosx)
= 1/(sinx cosx)
q.e.d.
Sure, here's a clownish proof for you:
First, let's rewrite the left side of the equation as 2csc(2x) = 2/sin(2x).
Now, let's rewrite the right side of the equation as csc(x)^2tan(x) = (1/sin(x))^2 * sin(x)/cos(x).
Now, let's simplify the right side: (1/sin(x))^2 * sin(x)/cos(x) = sin(x)/sin(x)^2 * sin(x)/cos(x).
Since sin(x)/sin(x)^2 simplifies to 1/sin(x), we can rewrite the right side as 1/sin(x) * sin(x)/cos(x).
Now, let's simplify it even further: 1/sin(x) * sin(x)/cos(x) = 1/cos(x).
So, we have 2/sin(2x) = 1/cos(x).
To make things even more fun, let's convert sin(2x) to its double-angle identity: sin(2x) = 2sin(x)cos(x).
Now, let's substitute that back into the equation: 2/sin(2x) = 2/(2sin(x)cos(x)).
Since the 2's in the numerator and denominator cancel out, we're left with 1/(sin(x)cos(x)) = 1/cos(x).
And since 1/(sin(x)cos(x)) = 1/cos(x), our equation holds true!
So, even though this proof may seem like clowning around, we were able to find an equivalent expression for the left and right sides of the equation.
To prove the given equation, we'll start from the left-hand side (LHS) and simplify it step-by-step until it matches the right-hand side (RHS) of the equation.
LHS: 2csc(2x)
Recall that csc(x) is the reciprocal of sin(x):
LHS: 2 * (1 / sin(2x))
Using the double angle identity for sine, sin(2x) = 2sin(x)cos(x):
LHS: 2 * (1 / (2sin(x)cos(x)))
Canceling the factor of 2:
LHS: 1 / (sin(x)cos(x))
Recall that tan(x) is the ratio of sin(x) to cos(x):
LHS: 1 / (sin(x)cos(x)) = 1 / (tan(x)cos(x))
Using the reciprocal identity for tangent, 1 / tan(x) = cot(x):
LHS: 1 / (tan(x)cos(x)) = 1 / (cot(x)cos(x))
Recall that cot(x) is the reciprocal of tan(x):
LHS: 1 / (cot(x)cos(x)) = cos(x) / (cos(x)sin(x))
Canceling out the common factor of cos(x):
LHS: cos(x) / (cos(x)sin(x)) = 1 / sin(x)
Recall that csc(x) is the reciprocal of sin(x):
LHS: 1 / sin(x) = csc(x)
Therefore, the LHS of the equation, 2csc(2x), simplifies to csc(x), which matches the RHS of the equation, csc(x)^2tan(x). Hence, the given equation is proven.
To prove the equation 2csc(2x) = csc(x)^2 * tan(x), we can start by using trigonometric identities to simplify both sides of the equation.
Starting with the left-hand side (LHS):
LHS = 2csc(2x)
Using the reciprocal identity, we can rewrite csc(2x) as 1/sin(2x):
LHS = 2 * (1/sin(2x))
Now we can simplify it further by multiplying 2 to the denominator:
LHS = 2/(2sin(x)cos(x))
Simplifying it further, we will divide both the numerator and denominator by 2:
LHS = 1/(sin(x)cos(x))
Now let's simplify the right-hand side (RHS):
RHS = csc(x)^2 * tan(x)
Using the reciprocal identity for csc(x), we can rewrite csc(x) as 1/sin(x):
RHS = (1/sin(x))^2 * tan(x)
Squaring the reciprocal of sin(x) gives us:
RHS = (1/sin(x))^2 * tan(x) = (1/sin^2(x)) * tan(x) = tan(x)/sin^2(x)
Now let's simplify the right-hand side further by expressing tan(x) in terms of sin(x) and cos(x).
Using the identity tan(x) = sin(x)/cos(x), we can rewrite the RHS as:
RHS = (sin(x)/cos(x)) / sin^2(x) = sin(x) / (cos(x) * sin^2(x))
Using the identity sin^2(x) = 1 - cos^2(x), we can rewrite the RHS as:
RHS = sin(x) / (cos(x) * (1 - cos^2(x)))
Now let's simplify the right-hand side further by factoring out sin(x) in the denominator:
RHS = sin(x) / (cos(x) - cos^3(x))
Now let's simplify the right-hand side by dividing both the numerator and denominator by cos(x):
RHS = sin(x) / cos(x) - cos^3(x) / cos(x)
The first term sin(x) / cos(x) is equal to tan(x), so the equation becomes:
RHS = tan(x) - cos^2(x)
Therefore, we have:
RHS = tan(x) - cos^2(x)
Comparing the RHS with the simplified LHS, we see that they are equal:
LHS = 1/(sin(x)cos(x)) = tan(x) - cos^2(x) = RHS
Thus, we have successfully proven the equation 2csc(2x) = csc(x)^2 * tan(x).