Maths Question...

[Keen_Interest]

Please only answer if you actually know for sure...

 

Is:    [ 7Cos(t)-7Cos(7t) ] / [ 7Sin(7t)-7Sin(t) ]    Equivalent to: [ Cos(t)-Cos(7t) ] / [ Sin(7t)-Sin(t) ]

 

Thanks in advance! :)

[0s4x]

No it isnt. I dont know the exact value of each equation, but they are not the same thing

[Keen_Interest]

No it isnt. I dont know the exact value of each equation, but they are not the same thing

Tyvm. I had a feeling they weren't, but wasn't sure. :)

[AroundWayOtherThe]

Don't you just have to take out the 7/7 in the first equation to reach the second one? Am I missing painfully obvious here? Because as far as I can see, yes they are equivalent...

[0s4x]

Tyvm. I had a feeling they weren't, but wasn't sure. :)

 

Well the first equation has sevens in front of each term, so that answer has to be larger than the one without such things in it.

[Lukiose]

From how i see it,  [ 7Cos(t)-7Cos(7t) ] / [ 7Sin(7t)-7Sin(t) ] can become

7[ Cos(t)-Cos(7t) ] / 7[ Sin(7t)-Sin(t) ] from factorisation which ends up being

7/7 [ [Cos(t)-Cos(7t)] / [sin(7t)-Sin(t)] ]

= [ Cos(t)-Cos(7t) ] / [ Sin(7t)-Sin(t) ]

They are equivalent, someone failed his maths lol Not the TS :P

[Keen_Interest]

So they are equivalent? Basically, a mark scheme has confused me. SO! I'm looking for clarification.. 

[KaareKanin]

From how i see it,  [ 7Cos(t)-7Cos(7t) ] / [ 7Sin(7t)-7Sin(t) ] can become

7[ Cos(t)-Cos(7t) ] / 7[ Sin(7t)-Sin(t) ] which ends up being

7/7 [ [Cos(t)-Cos(7t)] / [sin(7t)-Sin(t)] ]

= [ Cos(t)-Cos(7t) ] / [ Sin(7t)-Sin(t) ]

They are equivalent

This

[KaareKanin]

So they are equivalent? Basically, a mark scheme has confused me. SO! I'm looking for clarification..

If you want, you can try this out by assigning some (different) random value to both cos(t) sin(t) cos(7t) and sin (7t) and see if you get the same result :-p

[Keen_Interest]

Thanks everyone!