Asked by Beatriz
Prove:
2cos^2(x)-1 = 1-tan^2(x)/ 1+tan^2(x)
Thank you!
2cos^2(x)-1 = 1-tan^2(x)/ 1+tan^2(x)
Thank you!
Answers
Answered by
MathMate
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?
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?
Answered by
Beatriz
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
Answered by
MathMate
Try using sin²(x) = 1 - cos²(x) to get your final answer.
Answered by
Beatriz
I forgot to come on here and post that I did eventually come up with an answer. Thank you for all your help.
Answered by
MathMate
Glad that you got it!
You're welcome!
You're welcome!
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