Question

2. Prove the trig identity.

sin x ( 1 + cot^2x)

I have a calculus exam tomorrow and have no idea how to solve this equation, please provide a detailed explanation. I'm slow.

Answer 1

To prove the identity sin x (1 + cot^2 x), we start with the right-hand side of the identity and try to simplify it using trigonometric identities:

sin x (1 + cot^2 x)
= sin x (1 + cos^2 x / sin^2 x) (since cot x = cos x / sin x)
= sin x (sin^2 x / sin^2 x + cos^2 x / sin^2 x) (adding fractions with a common denominator)
= sin x [(sin^2 x + cos^2 x) / sin^2 x] (combining the fractions inside the brackets using the identity sin^2 x + cos^2 x = 1)
= sin x (1 / sin^2 x) (since sin^2 x + cos^2 x = 1)
= sin x / sin^2 x
= csc x

Therefore, we have shown that sin x (1 + cot^2 x) = csc x, and the identity is proved.

Answer 2

We'll start with the left-hand side (LHS) of the identity, which is:

sin x (1 + cot^2 x)

We can rewrite cot^2 x as (cos x / sin x)^2, since cotangent is the reciprocal of tangent. Substituting this into the LHS, we get:

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

Now we can simplify the expression in the parentheses by using the identity:

tan^2 x + 1 = sec^2 x

Rearranging this identity, we get:

tan^2 x = sec^2 x - 1

Substituting this into our expression, we get:

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

= sin x (sin^2 x / sin^2 x + cos^2 x / sin^2 x)

= sin x ((sin^2 x + cos^2 x) / sin^2 x)

= sin x (1 / sin^2 x)

= sin x / sin^2 x

= 1 / sin x

Now we'll simplify the right-hand side (RHS) of the identity, which is:

csc x

We know that csc x is the reciprocal of sin x, so we can rewrite the RHS as:

1 / sin x

This is the same as the expression we obtained for the LHS, so we have shown that:

sin x (1 + cot^2 x) = csc x

And this proves the identity!

Remember, when you're proving trigonometric identities, it's important to be familiar with the fundamental trigonometric identities and the basic algebraic rules of manipulating equations

good luck with your exam