Trig Proof and Solving Algebraically

FLorrum

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1. sin^2x cos^5x = (sin^2x - 2sin^4x + sin^6x) cos x -I tried to foil out the right side, but got a ton of sincos terms that I'm not sure what to do with. My teacher said it could be easier to use the left side, but I have no idea how to get to the right side from the left.

2. Solve Algebraically, 4sinx tan - 3tan x + 20 sinx -15 -My teacher left a hint, which is factor. I would normally use the quadratic formula but since there is sin and tan terms I can't do that. I don't think that there are any common terms to factor out? I need help getting started with this one.

Thank You
 
I apologize because trig identity proofs were always my least favorite part of Precalculus, so I'm no help on number one. However, I believe I can give you a nudge in the right direction for number two.

Remember that \(\displaystyle tan x = \frac{sinx}{cosx} = sinx* \frac{1}{cosx} = sinx * secx \), so then you really have:

\(\displaystyle 4sinx\:sinx\:secx-3sinx\:secx+20sinx-15\)

And you can, indeed, factor out a common term of sin(x). Hope that helps.
 
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1. sin^2x cos^5x = (sin^2x - 2sin^4x + sin^6x) cos x -I tried to foil out the right side, but got a ton of sincos terms that I'm not sure what to do with. My teacher said it could be easier to use the left side, but I have no idea how to get to the right side from the left.

2. Solve Algebraically, 4sinx tan - 3tan x + 20 sinx -15 -My teacher left a hint, which is factor. I would normally use the quadratic formula but since there is sin and tan terms I can't do that. I don't think that there are any common terms to factor out? I need help getting started with this one.

Thank You

sin2x * cos5x

= sin2x * cos(x) * cos4x

= sin2x * cos(x) * [cos2x]2

= sin2x * cos(x) * [1 - sin2x]2

Now continue....
 
1. sin^2x cos^5x = (sin^2x - 2sin^4x + sin^6x) cos x -I tried to foil out the right side, but got a ton of sincos terms that I'm not sure what to do with. My teacher said it could be easier to use the left side, but I have no idea how to get to the right side from the left.

2. Solve Algebraically, 4sinx tan - 3tan x + 20 sinx -15 -My teacher left a hint, which is factor. I would normally use the quadratic formula but since there is sin and tan terms I can't do that. I don't think that there are any common terms to factor out? I need help getting started with this one.

Thank You
You were given a great hint by Subhotosh for number 1 but I do want to comment on what you said. Why would you foil out when it is very clear that you can factor out a sin^2x from the first factor on the right hand side AND that is one of the factors on the left hand side? This reduces the problem to showing that cos^5x=(1-2sin^2x+sin^4x)cosx or simply cos^4x=(1-2sin^2x+sin^4x)
 
1. sin^2x cos^5x = (sin^2x - 2sin^4x + sin^6x) cos x -I tried to foil out the right side, but got a ton of sincos terms that I'm not sure what to do with. My teacher said it could be easier to use the left side, but I have no idea how to get to the right side from the left.

2. Solve Algebraically, 4sinx tan - 3tan x + 20 sinx -15 -My teacher le ft a hint, which is factor. I would normally use the quadratic formula but since there is sin and tan terms I can't do that. I don't think that there are any common terms to factor out? I need help getting started with this one.

Thank You
Hint for 2: You can factor by grouping 4sinx tanx - 3tan x + 20 sinx -15= [4sinx tanx - 3tan x] + [20 sinx -15] = tanx[4sinx-3] + 5[4sinx-3] = ...
BTW, if you want to solve you need to have an equation. Maybe 4sinx tan - 3tan x + 20 sinx -15 = 0??
 
I apologize because trig identity proofs were always my least favorite part of Precalculus, so I'm no help on number one. However, I believe I can give you a nudge in the right direction for number two.

Remember that \(\displaystyle tan x = \frac{sinx}{cosx} = sinx* \frac{1}{cosx} = sinx * secx \), so then you really have:

\(\displaystyle 4sinx\:sinx\:secx-3sinx\:secx+20sinx-15\)

And you can, indeed, factor out a common term of sin(x). Hope that helps.
ksdhart, sorry but sinx is NOT a common factor as the last term, 15, does not have a factor of sinx. Nice try!
 
ksdhart, sorry but sinx is NOT a common factor as the last term, 15, does not have a factor of sinx. Nice try!

Oh, dear. You're absolutely right. Sorry about any confusion I may have caused with that. My mistake seems so glaringly obvious now.
 
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