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C3 Trig Fraction - why do they ignore the denominator? Watch

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    Why do they just completely ignore the denominator? They replace 2sin^2x with 2(1-cos^2x) but they seem to ignore the sinxcosx denominator? Why is this? I'm really confused.

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    (Original post by vector12)
    Why do they just completely ignore the denominator? They replace 2sin^2x with 2(1-cos^2x) but they seem to ignore the sinxcosx denominator? Why is this? I'm really confused.
    The denom isn't very useful and can be cancelled.
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    (Original post by vector12)
    Why do they just completely ignore the denominator? They replace 2sin^2x with 2(1-cos^2x) but they seem to ignore the sinxcosx denominator? Why is this? I'm really confused.

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    Think of it as if the denominators where just numbers...
    if you had an expression over 5 equal to another expression over 5 you would just cancel them both so because t makes no difference, you can do the same thing here

    if you multiply both sides by the denominator then it disappears which is essentially what they've done
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    (Original post by Canine Copycat)
    Think of it as if the denominators where just numbers...
    if you had an expression over 5 equal to another expression over 5 you would just cancel them both so because t makes no difference, you can do the same thing here

    if you multiply both sides by the denominator then it disappears which is essentially what they've done
    That makes sense when you think of it like that involving an equals sign, like 2/5 = x/5, then x=5.

    However, I did it like this where I moved it all to the LHS and made it equal to 0, which doesn't seem right to cancel the denominator that way, since you can only do it if something equals something else, over the same denominator. I guess I just needed to realise that when they were equal, i could cancel the denominator?

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    (Original post by vector12)
    That makes sense when you think of it like that involving an equals sign, like 2/5 = x/5, then x=5.

    However, I did it like this where I moved it all to the LHS and made it equal to 0, which doesn't seem right to cancel the denominator that way, since you can only do it if something equals something else, over the same denominator. I guess I just needed to realise that when they were equal, i could cancel the denominator?
    You just need to realise that \displaystyle \frac{f(x)}{h(x)}=\frac{g(x)}{h(  x)} \Rightarrow f(x)=g(x) by simply multiplying both sides of the equation by h(x).

    Or, like you did, get \frac{f(x)-g(x)}{h(x)}=0 and then mult both sides by h(x) - but this is a redundant step as you can observe, since you still get f(x)-g(x)=0 \Rightarrow f(x)=g(x)
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    if you have

    x/5 = w/5

    then you can say

    x = w
 
 
 
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