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myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1well 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)

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1let me know if you still don't know where to go after that

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1do you know hot to factor the expression ax+x?

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1I 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 \[ax+1x=x(a+1)\]

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

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

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1also this way i'm asking you to approach it is not the only approach to take

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1ok 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?

myininaya
 6 months ago
Best 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?

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1right so what does that leave us with?

myininaya
 6 months ago
Best 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?

kengeta
 6 months ago
Best ResponseYou've already chosen the best response.0sin^2(theta)+cos^2(theta)

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1you need to recall some trig identities for these trigonometric proofs

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1This one of the most basic ones It actually has a name

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1I will give you a hint: Pythagorean Identity

kengeta
 6 months ago
Best ResponseYou've already chosen the best response.0okay i get it now thanks soooo much

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1Way 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.

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1You don't have to do both ways. Just one way.

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1I used a Pythagorean identity in both ways.

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1way 1 was the way I was trying to get you to go about it the first time around

myininaya
 6 months ago
Best ResponseYou've already chosen the best response.1I 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.
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