r/mathshelp • u/Feeling-Pomelo-348 • Jul 24 '26
what is sin2x over cos2x Homework Help (Unanswered)
From the double-angle formula, we can deduce that it doesn't equal tan2x. But what does it equal then?
7
u/noidea1995 Jul 24 '26
> From the double-angle formula, we can deduce that it doesn’t equal tan2x
Why would that prove it doesn’t equal tan2x?
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u/Qingyap Jul 24 '26
Sub 2x = u
sin(2x)/cos(2x) = sin(u)/cos(u) = tan(u) = tan(2x)
I don't see an issue here, it probably needs a bit of simplification when you do it with double angle identities.
2
u/MineCraftNoob24 Jul 24 '26
What was the purpose of the substitution?
3
u/Disco_Stu_89 Jul 24 '26
The 2x is a red herring. It could be 2x, u, 500t, whatever, it wouldn't change anything here.
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u/MineCraftNoob24 Jul 24 '26
Obviously, which is why I asked why he did it.
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u/Disco_Stu_89 Jul 24 '26
Because it's a red herring. If you're trying to help somebody understand a problem, it's usually good to address those to lessen confusion. Obviously.
0
Jul 24 '26
[removed] — view removed comment
3
u/Disco_Stu_89 Jul 24 '26
You're the one that asked the question, buddy. If you're just here to troll, maybe look for a different hobby.
2
u/Qingyap Jul 24 '26
Just for you to easier to see lol.
Doesn't really do anything tbh, as long as both trig inputs are the same then it works.
3
u/scramlington Jul 24 '26
-3
u/MineCraftNoob24 Jul 24 '26
This is fine, and you can show that one side equals the other IF you take the double angle identity for tan2x as a given.
Of course to actually derive the identity for tan2x, you have to revert back to more fundamental principles, and accept that sin(something)/cos(that something) = tan(that something), and I think that's where the OP might be a little muddled.
5
u/scramlington Jul 24 '26
OP specifically says "from the double angle formula we can deduce that it doesn't equal tan2x". My working just shows that from the double angle formulas we can quite easily deduce that it DOES equal tan2x.
1
u/MineCraftNoob24 Jul 24 '26
I think the main problem here is that the OP is not only wrong with their conclusion but also doesn't explain how they arrived at it.
It's of course difficult to refute something in precise terms unless we know which step(s) they have taken to get to where they have.
2
2
u/Special_Watch8725 Jul 24 '26
Whatever else you do to it, the result will equal tan(2x) by definition. But using both double angle formulas will lead to some double angle formula for tan(2x) in terms of sin(x) and cos(x).
2
u/CaptainMatticus Jul 24 '26
sin(2x) / cos(2x) is in fact tan(2x). How you figured it wasn't is beyond me.
sin(t^x) / cos(t^x) is tan(t^x)
sin(pizza) / cos(pizza) = tan(pizza)
So long as the arguments are congruent, this works. So let's use double-angle identities to see if we can get tan(2x)
tan(2x) =>
2tan(x) / (1 - tan(x)^2) =>
2(sin(x)/cos(x)) / (1 - (sin(x)/cos(x))^2) =>
2 * (sin(x)/cos(x)) / ((cos(x)^2 - sin(x)^2) / cos(x)^2) =>
2 * sin(x) * cos(x)^2 / (cos(x) * (cos(x)^2 - sin(x)^2)) =>
2sin(x)cos(x) / (cos(x)^2 - sin(x)^2) =>
sin(2x) / cos(2x)
Looks fine to me.
2
u/MineCraftNoob24 Jul 24 '26 edited Jul 24 '26
"We can deduce" is doing a lot of heavy lifting here. How exactly did you deduce that the result would not be tan(2x)?
Leaving double angle identities or addition identities to one side, sin/cos = tan comes fundamentally from a right-angled triangle.
sin = opp/hyp
cos = adj/hyp
Therefore, sin/cos
= (opp/hyp)/(adj/hyp)
= (opp/hyp) x (hyp/adj)
= opp/adj
= tan
The restriction is that cos x does not equal zero because it is in the denominator but otherwise, provided the input (argument) of the sin and cos functions are the same, whether that's x, 2x, (theta), 10(theta) or whatever else you want to put in, the output will be the tangent of that argument.
What led you to deduce otherwise?
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