Prove:
2cos^2(x)-1 = 1-tan^2(x)/ 1+tan^2(x)
Thank you!
Examine both sides of the equation.
The right-hand side can be reduced to sin(x) and cos(x), as is most cases involving tangents.
So start from the right-hand side:
(1-tan²(x))/(1+tan²(x))
=(((cos²(x)-sin²(x))/cos²(x)) / (((cos²(x)+sin²(x))/cos²(x))
Cancel out the cos²(x) and substituting 1 for (cos²(x)+sin²(x)) leaves us with
((cos²(x)-sin²(x))/1
=((cos²(x)-sin²(x))
Can you continue from here?
Thank you so much for your help. I followed it and understood it, yet I do not know what to do afterwards. If you could still help me that would be great. Thank you! =D
Try using sin²(x) = 1 - cos²(x) to get your final answer.
I forgot to come on here and post that I did eventually come up with an answer. Thank you for all your help.
Glad that you got it!
You're welcome!
To prove the given equation, we need to simplify the left-hand side (LHS) and the right-hand side (RHS) separately and show that they are equal.
Starting with the LHS:
2cos^2(x) - 1
Using the identity cos^2(x) = 1 - sin^2(x), we can rewrite this as:
2(1 - sin^2(x)) - 1
Simplifying further:
2 - 2sin^2(x) - 1
Combining like terms:
1 - 2sin^2(x)
Moving on to the RHS:
1 - tan^2(x) / (1 + tan^2(x))
Using the identity tan^2(x) = sin^2(x) / cos^2(x), we can substitute this into the RHS:
1 - (sin^2(x) / cos^2(x)) / (1 + (sin^2(x) / cos^2(x)))
Simplifying the denominator:
1 - sin^2(x) / cos^2(x) / (1 + sin^2(x) / cos^2(x))
Now we can simplify the numerator:
1 - sin^2(x) / cos^2(x) / (cos^2(x) + sin^2(x)) / cos^2(x)
Since cos^2(x) + sin^2(x) = 1, we can substitute this into the RHS:
1 - sin^2(x) / cos^2(x) / (1 / cos^2(x))
Now, inverting the denominator:
1 - sin^2(x) / cos^2(x) * cos^2(x) / 1
Simplifying:
1 - sin^2(x)
We can see that the simplified equations for both the LHS and the RHS are the same: 1 - 2sin^2(x) = 1 - sin^2(x).
Thus, we have proven that 2cos^2(x) - 1 = 1 - tan^2(x) / (1 + tan^2(x)).