Trigonometry identities are so hard...
I need some help proving these identities:
*Oh, and I'm only in grade 11, so the identities we use are quotient identity and Pythagorean identity.
sinx/(sinx + cosx) = tanx/(1 + tanx)
cos^2x - sin^2x = 2cos^2x - 1
Thanks!
Lucy
By inverting the fractions (a perfectly legal operation), the first equation can be converted to
(sin x + cos x)/sin x = (1 + tan x)/x
1 + cot x = 1 + 1/tan x
= 1 + cot x
In the second problem, substitute 1 - cos^2 x for sin^2 x on the left side.
Thanks for your help!
I understand the second problem now.
Except I'm confused about what you did in the first problem. We haven't learned anything about cotx yet...
I inverted the fractions, though and ended up with:
(sin x + cos x) / sin x = (1 + tan x) / tanx
(I'm just wondering...why did you write (1 + tan x) / x on the left side?)
Then simplifies to... cos x = 1 ??
I'm confused... :S
*Sorry, should be:
I'm just wondering...why did you write (1 + tan x) / x on the *right* side?
Instead of (1 + tan x) / tan x?
SOH CAH TOA... IT'S sin=opposite over hypotenus..... Cos=ajacent over hypotenus.... Tan=opposite over ajacent
SOH CAH TOA... IT'S sin=opposite over hypotenus..... Cos=ajacent over hypotenus.... Tan=opposite over ajacent..... When you want to fine thiter or any angle
Hi Lucy! Trigonometric identities can be challenging, but don't worry, I'll do my best to help you understand how to prove these identities step by step.
Let's start with the first identity:
sinx / (sinx + cosx) = tanx / (1 + tanx)
To prove this identity, we'll work on the left-hand side (LHS) and the right-hand side (RHS) separately and verify that they are equal.
Starting with the LHS:
sinx / (sinx + cosx)
To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which in this case is (sinx - cosx). This is a common technique used in trigonometric proofs.
(sin x / (sin x + cos x)) * ((sin x - cos x) / (sin x - cos x))
Now, let's distribute and simplify the numerator:
sin^2x - sinx*cosx
Now, moving on to the RHS:
tanx / (1 + tanx)
To simplify this expression, we can replace tanx with sinx/cosx since tanx is equivalent to sinx/cosx.
(sin x / cos x) / (1 + (sin x / cos x))
Next, we simplify the numerator:
sin x / cos x
Now, let's find a common denominator for the denominator in order to simplify further:
(cos x / cos x) + (sin x / cos x)
Simplifying the denominator:
(1 + sin x) / cos x
Now, comparing the simplified numerator and denominator of the LHS with the RHS, we can see that they are equivalent. Thus, the LHS = RHS, and the identity is proven.
Now let's move on to the second identity:
cos^2x - sin^2x = 2cos^2x - 1
We'll start with the LHS:
cos^2x - sin^2x
Using the Pythagorean identity sin^2x + cos^2x = 1, we can replace sin^2x with 1 - cos^2x:
cos^2x - (1 - cos^2x)
Next, let's distribute the negative sign and combine like terms:
2cos^2x - 1
As we can see, the simplified LHS is equal to the RHS, so the identity is proven.
I hope this helps you understand how to prove these identities using the quotient and Pythagorean identities. Feel free to ask if you have any more questions!