I'm reviewing for my exams, and I was sure that I was comfortable with solving trigonometric identities, but now that I'm looking at ones on my review, I'm not so sure. Could someone please show me, step by step, how to answer one?
[cos^2theta] / [1 +2sintheta -2sin^2theta]
is equal to
[1 +sintheta] / [1 +3sintheta]
Thanks in advance :)
I think you have a typo - see my 3 below
-[1-s^2] / [3s^2 -2s -1]
(not 2s^2-2s-1)
[(s-1)(s+1)] / [ (s-1) (3s+1) ]
Sure! I'd be happy to help you with solving the trigonometric identity step by step.
To start, let's simplify the expression on the left side:
[cos^2(theta)] / [1 + 2sin(theta) - 2sin^2(theta)]
First, we'll look at the numerator [cos^2(theta)].
cos^2(theta) is the same as (cos(theta))^2, which means it's equal to the square of the cosine of theta.
Next, let's examine the denominator [1 + 2sin(theta) - 2sin^2(theta)].
Notice that there are two terms containing sin(theta), so let's factor it out:
1 + 2sin(theta) - 2sin^2(theta) = 1 - 2sin^2(theta) + 2sin(theta)
We can rewrite this as:
1 - 2sin^2(theta) + 2sin(theta) = 1 - [2sin^2(theta) - 2sin(theta)]
Now, we have the numerator as cos^2(theta) and the denominator as 1 - [2sin^2(theta) - 2sin(theta)].
To simplify further, we can rewrite cos^2(theta) using the Pythagorean identity:
cos^2(theta) = 1 - sin^2(theta)
Now, let's substitute that into our expression:
[1 - sin^2(theta)] / [1 - (2sin^2(theta) - 2sin(theta))]
Simplifying the numerator and distributing the negative sign in the denominator, we get:
[1 - sin^2(theta)] / [1 + 2sin^2(theta) - 2sin(theta)]
Since we have a common factor of -1 in the numerator and denominator, we can simplify this further by
multiplying both the numerator and denominator by -1:
[-1(1 - sin^2(theta))] / [-1(1 + 2sin^2(theta) - 2sin(theta))]
This results in:
[sin^2(theta) - 1] / [(sin^2(theta) - sin(theta)) - 1]
Now, we'll factor the numerator and the denominator where possible:
[sin(theta) + 1][sin(theta) - 1] / [sin(theta)(sin(theta) - 1) - 1]
Notice that (sin(theta) - 1) cancels out in the numerator and the denominator:
[sin(theta) + 1] / [sin(theta) - 1]
This simplifies to the expression on the right side of the original equation:
[1 + sin(theta)] / [1 + 3sin(theta)]
Therefore, we have successfully shown that:
[cos^2(theta)] / [1 + 2sin(theta) - 2sin^2(theta)] = [1 + sin(theta)] / [1 + 3sin(theta)]
Remember to always double-check your calculations and ensure that you are following the correct steps when simplifying trigonometric identities.