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myininaya Group TitleBest ResponseYou've already chosen the best response.1
well my first attempt to prove this is true is to take the left hand side and try to show the right hand side the left hand side has both terms with the factor cos^2(theta) start by factoring cos^2(theta) out from both terms this will get us closer to writting as 1 term just as the right hand side is (one term)
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
let me know if you still don't know where to go after that
 one month ago

kengeta Group TitleBest ResponseYou've already chosen the best response.0
I'm pretty confused
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
on how to factor?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
do you know hot to factor the expression ax+x?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
how* (not hot)
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I will show you how to factor ax+x then you should be able to factor tan^2(theta)*cos^2(theta)+cos^2(theta) so ax+x I see there is a x in both terms dw:1407522006591:dw so i will factor that x out like so \[ax+1x=x(a+1)\]
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you have \[\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1 \] \[\text{ do you see that there is a } cos^2(\theta) \text{ in both terms ?}\]
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
can you factor that out using what you know about multiplication and division like i did above?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
also this way i'm asking you to approach it is not the only approach to take
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
ok if you are having trouble factoring the cos^2(theta) out how about we try this another way do you recall tan is sin/cos?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta)\] in the first term there, do you see anything that cancels?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
right so what does that leave us with?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\sin^2(\theta)}{\cancel{\cos^2(\theta)}}\cancel{\cos^2(\theta)}+\cos^2(\theta)\] what does this give us?
 one month ago

kengeta Group TitleBest ResponseYou've already chosen the best response.0
sin^2(theta)+cos^2(theta)
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
And that equals?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you need to recall some trig identities for these trigonometric proofs
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
This one of the most basic ones It actually has a name
 one month ago

kengeta Group TitleBest ResponseYou've already chosen the best response.0
okay si... 1?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I will give you a hint: Pythagorean Identity
 one month ago

kengeta Group TitleBest ResponseYou've already chosen the best response.0
okay i get it now thanks soooo much
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Way 1: \[\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1= \\ \cos^2(\theta)(\tan^2(\theta)+1)= \\ \cos^2(\theta) \cdot \sec^2(\theta)= \\ 1\] Way 2: \[\tan^2(\theta) \cos^2(\theta)+\cos^2(\theta)= \\ \frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta) = \\ \sin^2(\theta) +\cos^2(\theta)= \\ 1\] These are the two ways that I can think of... This does not mean they are the only ways.
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You don't have to do both ways. Just one way.
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I used a Pythagorean identity in both ways.
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
way 1 was the way I was trying to get you to go about it the first time around
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I don't think one way is more harder than the other But you will have to review some algebra especially if you don't remember how to factor. Factoring will come up again.
 one month ago
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