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The graph of tan(sinθ)−sin(tanθ) is given below:
alt text
This provides a lot of information, for example, that tan(sinθ)−sin(tanθ) is not an increasing function. Furthermore, your explanation must be able to provide a reason why the value approach back to 0 as θ approaches π.
As we know from looking at the graph of the sinθ and tanθ functions, we know that the tanθ graph starts increasing at an increasing rate after θ=0 and the sinθ graph starts increasing at a decreasing rate. At θ=0, they are both equal. This only applies for 0<θ<2π, because from 2π<θ<π, tanθ becomes negative. now that we have established that, we should know that from about 0<θ≤4π, we can take tanθ≈sinθ. Therefore sin(tanθ) will be lesser than tan(sinθ), as tanθ is greater than sinθ for 0<θ<2π.
from 4π<θ<2π, tanθ begins to become much greater than sinθ. Despite that, sin(tanθ) can only have values for −1≤sin(tanθ)≤1. From 4π<θ≤2π, the minimum value of sinθ is 22 and the maximum is 1. hence the minimum for tan(sinθ is 0.854 and 1.56. since we know that tan1≈2π, we know that 4π≈1, we can deduce that sin(tanθ) will decrease from its maximum of 1, so tan(sinθ) can increase from its minimum of 0.854, and it can increase beyond 1, and therefore pass the reach of sin(tanθ.
Since trigonometric graphs are periodical, tan(sinθ) is also periodical, and knowing it hits its maximum at θ=2π, it must go back to 0 at θ=π. since tan(sinθ) started its period at the same time as sin(tanθ), then both of them will hit 0 θ=π. However using differentiation, we can deduce that tan(sinθ) is decreasing at θ=π, whereas sin(tanθ) is increasing. Hence, sin(tanθ)<tan(sinθ) from 0<θ<π
This is a good start, with several relevant ideas. However, you need to be really careful with your arguments. They are currently too 'handwavy' and do not provide any proper arguments.
In the first paragraph, if all we know is that tanθ≈sinθ, how does that give us the result? You cannot say that sin(tanθ)≈sin(sinθ)<tan(sinθ), because in the first 'approximation', we actually have sin(tanθ)>sin(sinθ).
In your second paragraph, you seem to be arguing that for 4π<θ<2π, we have tan(sinθ)≥0.854 (approx). However, this is not enough to conclude that tan(sinθ)≥1≥sin(tanθ), which seems to be your argument. Note that sin(tanθ) in the range 4π<θ<2π is a highly oscillating function, which explains the middle portion of the graph that I attached above. (This occurs because sin is periodic, and tanθ goes towards infinity. It is not a constantly decreasing function.
The difficulty caused by oscillating sin also extends to the range 2π<θ<43π, as tanθ increases from −∞. The argument that you provided doesn't hold over the rest of the range, but could hold over a reduced range. In particular, you should carry out the differentiation and state the ranges in which your conditions hold, and ensure that it overlaps with the reduced range.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
a_{i-1}
\frac{2}{3}
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\sum_{i=1}^3
\sin \theta
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Comments
The graph of tan(sinθ)−sin(tanθ) is given below:
alt text
This provides a lot of information, for example, that tan(sinθ)−sin(tanθ) is not an increasing function. Furthermore, your explanation must be able to provide a reason why the value approach back to 0 as θ approaches π.
But tanx is undefined at x=2π
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But still taking x=2π we get
sin (tan 2π) and tan (sin 2π)
Let tan 2π be α
As per range of sinθ it cannot exceed 1
so sinα < 1 or =1
But tan (sin 2π) = tan(1) > 1
And it is applicable for 2π
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You can't write anything involving tan2π, since it's undefined! Even though sinx≤1, you can't take the sine of an undefined number.
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As we know from looking at the graph of the sinθ and tanθ functions, we know that the tanθ graph starts increasing at an increasing rate after θ=0 and the sinθ graph starts increasing at a decreasing rate. At θ=0, they are both equal. This only applies for 0<θ<2π, because from 2π<θ<π, tanθ becomes negative. now that we have established that, we should know that from about 0<θ≤4π, we can take tanθ≈sinθ. Therefore sin(tanθ) will be lesser than tan(sinθ), as tanθ is greater than sinθ for 0<θ<2π.
from 4π<θ<2π, tanθ begins to become much greater than sinθ. Despite that, sin(tanθ) can only have values for −1≤sin(tanθ)≤1. From 4π<θ≤2π, the minimum value of sinθ is 22 and the maximum is 1. hence the minimum for tan(sinθ is 0.854 and 1.56. since we know that tan1≈2π, we know that 4π≈1, we can deduce that sin(tanθ) will decrease from its maximum of 1, so tan(sinθ) can increase from its minimum of 0.854, and it can increase beyond 1, and therefore pass the reach of sin(tanθ.
Since trigonometric graphs are periodical, tan(sinθ) is also periodical, and knowing it hits its maximum at θ=2π, it must go back to 0 at θ=π. since tan(sinθ) started its period at the same time as sin(tanθ), then both of them will hit 0 θ=π. However using differentiation, we can deduce that tan(sinθ) is decreasing at θ=π, whereas sin(tanθ) is increasing. Hence, sin(tanθ)<tan(sinθ) from 0<θ<π
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This is a good start, with several relevant ideas. However, you need to be really careful with your arguments. They are currently too 'handwavy' and do not provide any proper arguments.
In the first paragraph, if all we know is that tanθ≈sinθ, how does that give us the result? You cannot say that sin(tanθ)≈sin(sinθ)<tan(sinθ), because in the first 'approximation', we actually have sin(tanθ)>sin(sinθ).
In your second paragraph, you seem to be arguing that for 4π<θ<2π, we have tan(sinθ)≥0.854 (approx). However, this is not enough to conclude that tan(sinθ)≥1≥sin(tanθ), which seems to be your argument. Note that sin(tanθ) in the range 4π<θ<2π is a highly oscillating function, which explains the middle portion of the graph that I attached above. (This occurs because sin is periodic, and tanθ goes towards infinity. It is not a constantly decreasing function.
The difficulty caused by oscillating sin also extends to the range 2π<θ<43π, as tanθ increases from −∞. The argument that you provided doesn't hold over the rest of the range, but could hold over a reduced range. In particular, you should carry out the differentiation and state the ranges in which your conditions hold, and ensure that it overlaps with the reduced range.
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Thanks for your feedback!
Thats an nice attempt but can we not prove it by any other way I mean not by graphs .