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Note: This is not homework. This is from a monthly math challenge that my university offers. This particular question is past its due date, and I'm discovering it late. I'm an engineering student who's fascinated by math and wants to give problems from the math department a shot (though I have no idea if I even have the tools to solve some of these, as I've only just finished Calc II).
(a) Show that there exists exactly one real solution of [itex] sinx +sin^2x = 1[/itex] for [itex]
0 ≤ x ≤ π/2 [/itex]
(b) Show that for each n ∈ N there exists exactly one real solution of
[itex]sin x + (sin^n x) = 1[/itex] for [itex] 0 ≤ x ≤ π/2[/itex]
N/A
I've never really had to use proofs in my math classes so far. I assume I'll need the intermediate value theorem.
For part (a), we see that [itex]sin(0) + sin^2(0) = 0 < 1[/itex] and that
[itex]sin(π/2) + sin^2(π/2) = 1 + 1 = 2 > 1[/itex]. Since [itex]f(x) = sinx + sin^2x[/itex] is continuous for all x, and since [itex]f(0) < 1 < f(π/2)[/itex], there must exist some c ∈ [0, π/2] such that f(c) = 1. This would be good I believe if it weren't for the condition that there is exactly one real solution, which the intermediate value theorem wouldn't be able to let us know. Am I approaching this wrong? Help would be appreciated (as I don't have any background in writing proofs).
Part (b) I feel encounters the same difficulty.
Thanks.
Homework Statement
(a) Show that there exists exactly one real solution of [itex] sinx +sin^2x = 1[/itex] for [itex]
0 ≤ x ≤ π/2 [/itex]
(b) Show that for each n ∈ N there exists exactly one real solution of
[itex]sin x + (sin^n x) = 1[/itex] for [itex] 0 ≤ x ≤ π/2[/itex]
Homework Equations
N/A
The Attempt at a Solution
I've never really had to use proofs in my math classes so far. I assume I'll need the intermediate value theorem.
For part (a), we see that [itex]sin(0) + sin^2(0) = 0 < 1[/itex] and that
[itex]sin(π/2) + sin^2(π/2) = 1 + 1 = 2 > 1[/itex]. Since [itex]f(x) = sinx + sin^2x[/itex] is continuous for all x, and since [itex]f(0) < 1 < f(π/2)[/itex], there must exist some c ∈ [0, π/2] such that f(c) = 1. This would be good I believe if it weren't for the condition that there is exactly one real solution, which the intermediate value theorem wouldn't be able to let us know. Am I approaching this wrong? Help would be appreciated (as I don't have any background in writing proofs).
Part (b) I feel encounters the same difficulty.
Thanks.