Sin x/[1-cos(-x)] = csc x + cot x
Please Verify the Identity.
I have been going in circle with this one... Just a hint would be nice. Thank you.
LS = sinx/(1 - cos(-x))
first of all cos(-x) = cosx, so let's change that
= sinx/(1 - cosx)
now multiply top and bottom by (1+cosx)/(1+cos),
thus changing the appearance, but not its value
= sinx/(1-cosx) * (1+cosx)/(1+cos)
= sinx(1+cosx)/(1 - cos^2 x)
= (sinx + sinxcosx)/sin^2 x
= sinx/sin^2 x + sinxcosx/sin^2 x
= 1/sinx + cosx/sinx
= cscx + cotx
= RS
Thank you so much!
To verify the identity sin x/[1-cos(-x)] = csc x + cot x, we need to simplify both sides of the equation till they are equivalent.
Let's start with simplifying the left side of the equation:
sin x/[1-cos(-x)]
First, let's simplify the denominator by using the identity cos(-x) = cos x.
sin x/[1-cos x]
Next, let's simplify the left side further:
sin x/[1-cos x] = (sin x * (1+cos x))/[(1-cos x) * (1+cos x)]
Now, let's simplify the right side of the equation:
csc x + cot x
Recall that csc x = 1/sin x and cot x = cos x/sin x:
csc x + cot x = 1/sin x + cos x/sin x
Combining the fractions, we have:
csc x + cot x = (1 + cos x)/sin x
Now, if we compare the simplified left side and right side of the equation, we can see that they are equal:
[(sin x * (1+cos x))/[(1-cos x) * (1+cos x)]] = (1 + cos x)/sin x
To further simplify, we can cancel out the common factor of (1+cos x) in the numerator and denominator on the left side:
(sin x * (1+cos x))/[(1-cos x) * (1+cos x)] = sin x/(1-cos x)
Now, both sides are equal:
sin x/(1-cos x) = sin x/(1-cos x)
Therefore, the identity sin x/[1-cos(-x)] = csc x + cot x is verified.
To verify the given identity, we need to simplify both sides to see if they are equal.
Starting with the left-hand side (LHS):
LHS = sin(x) / [1 - cos(-x)]
We can simplify this expression by using the evenness property of cosine. The evenness property states that cos(-x) = cos(x). Therefore, we can rewrite the LHS as:
LHS = sin(x) / [1 - cos(x)]
Next, let's simplify the right-hand side (RHS):
RHS = csc(x) + cot(x)
Using the reciprocal identities, we can express cot(x) and csc(x) in terms of sine and cosine as follows:
RHS = 1/sin(x) + cos(x)/sin(x)
To simplify further, we need to find a common denominator for the two terms on the RHS:
RHS = (1 + cos(x)) / sin(x)
Now, let's compare the simplified LHS and RHS:
LHS = sin(x) / [1 - cos(x)]
RHS = (1 + cos(x)) / sin(x)
To determine if the given identity holds true, we need to check if LHS equals RHS. We can do this by cross-multiplying:
LHS = RHS
Cross-multiplying, we get:
sin(x) * sin(x) = [1 - cos(x)] * (1 + cos(x))
Expanding the right-hand side, we have:
sin^2(x) = 1 - cos^2(x)
Recalling the Pythagorean identity, sin^2(x) + cos^2(x) = 1, we can substitute this into the equation:
1 - cos^2(x) = 1 - cos^2(x)
Both sides of the equation are equal to each other, so the given identity is verified as true.
Therefore, sin(x) / [1 - cos(-x)] = csc(x) + cot(x) is an identity.