- #1
Snoozems
- 3
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Hi there,
This is my very first post, so I'd like to say thanks for reading and hi basically.
I'm relatively confident my attempt at the proof is correct, but since the method is quite different from other examples I have seen, it kind of makes me nervous. I was hoping someone could verify my answer?
Prove that:
[itex] \sin^2\theta (1 + sec^2\theta) = sec^2 \theta - cos^2 \theta[/itex]
So I try manipulate the right hand side to take exactly the same form as the left. First I rewrite the terms as their 'ordinary' equivalents:
##\dfrac{1}{\cos^2 \theta} - \cos^2 \theta##
##\dfrac{1}{\cos^2 \theta} + \sin^2 \theta - 1##
I then combine the trig terms:
##\dfrac{\sin^2 \theta \cos^2 \theta + 1}{\cos^2 \theta} - 1##
Now I assimilate the constant term of -1:
##\dfrac{\sin^2 \theta \cos^2 \theta - \cos^2\theta + 1}{\cos^2 \theta}##
I now use the fact that ##-\cos^2 \theta + 1 = \sin^2 \theta##:
##\dfrac{\sin^2 \theta \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}##
I remove the common factor of ##\sin^2 \theta## from the numerator:
##\dfrac{\sin^2 \theta( \cos^2 \theta + 1)}{\cos^2 \theta}##
Next I split the fraction into its constituent products:
##\sin^2 \theta \left(\dfrac{\cos^2 \theta + 1}{\cos^2 \theta} \right)##
Simplifying:
##\sin^2 \theta \left(1 + \dfrac{1}{\cos^2 \theta} \right) \equiv \sin^2\theta (1 + sec^2\theta)##
Have I gone wrong anywhere at all or does this work just fine? Thanks a lot!
~Snooz
This is my very first post, so I'd like to say thanks for reading and hi basically.
I'm relatively confident my attempt at the proof is correct, but since the method is quite different from other examples I have seen, it kind of makes me nervous. I was hoping someone could verify my answer?
Homework Statement
Prove that:
[itex] \sin^2\theta (1 + sec^2\theta) = sec^2 \theta - cos^2 \theta[/itex]
Homework Equations
The Attempt at a Solution
So I try manipulate the right hand side to take exactly the same form as the left. First I rewrite the terms as their 'ordinary' equivalents:
##\dfrac{1}{\cos^2 \theta} - \cos^2 \theta##
##\dfrac{1}{\cos^2 \theta} + \sin^2 \theta - 1##
I then combine the trig terms:
##\dfrac{\sin^2 \theta \cos^2 \theta + 1}{\cos^2 \theta} - 1##
Now I assimilate the constant term of -1:
##\dfrac{\sin^2 \theta \cos^2 \theta - \cos^2\theta + 1}{\cos^2 \theta}##
I now use the fact that ##-\cos^2 \theta + 1 = \sin^2 \theta##:
##\dfrac{\sin^2 \theta \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}##
I remove the common factor of ##\sin^2 \theta## from the numerator:
##\dfrac{\sin^2 \theta( \cos^2 \theta + 1)}{\cos^2 \theta}##
Next I split the fraction into its constituent products:
##\sin^2 \theta \left(\dfrac{\cos^2 \theta + 1}{\cos^2 \theta} \right)##
Simplifying:
##\sin^2 \theta \left(1 + \dfrac{1}{\cos^2 \theta} \right) \equiv \sin^2\theta (1 + sec^2\theta)##
Have I gone wrong anywhere at all or does this work just fine? Thanks a lot!
~Snooz