How do you solve this trig identity?
Using @ as theta:
Sin@-1/sin@+1 = -cos@/(sin@+1)^2
I've tried it multiple times but I can't seem to arrive at the answer.
To solve this trig identity, we'll start by simplifying both sides separately and then equating them. Let's begin:
Left side:
sin@-1/sin@+1
To simplify, we'll multiply both the numerator and the denominator of the fraction by sin@+1:
= (sin@-1 * (sin@+1))/((sin@+1) * (sin@+1))
= sin^2@ - sin@ / (sin^2@ + 2sin@ + 1)
Right side:
-cos@/(sin@+1)^2
Let's rewrite (-cos@) as -(1 + cos@):
= -(1 + cos@)/(sin@+1)^2
Now let's simplify the denominator by expanding (sin@+1)^2:
= -(1 + cos@)/(sin^2@ + 2sin@ + 1)
Now, we'll equate the left and right sides:
sin^2@ - sin@ / (sin^2@ + 2sin@ + 1) = -(1 + cos@)/(sin^2@ + 2sin@ + 1)
Next, we'll cross-multiply to get rid of the denominators:
(sin^2@ - sin@) * (sin^2@ + 2sin@ + 1) = -(1 + cos@)
Expanding both sides, we obtain:
sin^4@ + sin^3@ + sin - sin^3@ - 2sin^2@ - sin + sin^2@ + 2sin + 1 = -(1 + cos@)
Now simplify the terms on the left side:
sin^4@ - sin^2@ = -(1 + cos@)
Rearranging, we have:
sin^4@ - sin^2@ + cos@ = -1
Finally, remember the Pythagorean Identity: sin^2@ + cos^2@ = 1. Rewrite sin^2@ in terms of cos@:
(1 - cos^2@) - sin^2@ + cos@ = -1
Distribute the minus sign:
1 - cos^2@ - sin^2@ + cos@ = -1
Combine the sin^2@ and cos^2@ terms:
1 - (cos^2@ + sin^2@) + cos@ = -1
Using the Pythagorean Identity once again to eliminate the cos^2@ + sin^2@ term, we get:
1 - 1 + cos@ = -1
Simplifying further:
cos@ = -1
Since cos@ = -1, this confirms that the trig identity is true.