Challenging trigonometry problem
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mestmoha said:
\(\displaystyle (\sin A + \sin B \sin C)^2 \leq 1+ \sin A + \sin B \sin C\)
Start using,
\(\displaystyle A \ = \ \pi \ - \ (B \ + \ C)\)
Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
we should prove:
\(\displaystyle $\alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0$\)
we know that:
\(\displaystyle $\frac{\sin A}{a} = \frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$\)
such that R is the cimscribed circle radius
by Substituting in alpha, we need to prove now:
\(\displaystyle $\frac{16 R^4 + 8 aR^3 + 4 bc R^2 - (4 a^2 R^2 + 4 abc R + b^2 c^2) } {16 R^4} \geq 0 $\)
Such that a, b,and c are the edges of a triangle.
As a remark,
\(\displaystyle $R=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}$\)
Thank you,
\(\displaystyle $\alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0$\)
we know that:
\(\displaystyle $\frac{\sin A}{a} = \frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$\)
such that R is the cimscribed circle radius
by Substituting in alpha, we need to prove now:
\(\displaystyle $\frac{16 R^4 + 8 aR^3 + 4 bc R^2 - (4 a^2 R^2 + 4 abc R + b^2 c^2) } {16 R^4} \geq 0 $\)
Such that a, b,and c are the edges of a triangle.
As a remark,
\(\displaystyle $R=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}$\)
Thank you,
I think I found the solution. Please, check it for me:
\(\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)
\(\displaystyle \frac{{\sin}^2 A}{{a}^2}=\frac{\sin B \sin C}{bc}\)
\(\displaystyle \sin A + \sin B \sin C= \Sin A (1 + \frac{bc}{{a}^2} \sin A)\)
For any A, and a \(\displaystyle \sin A + \sin B \sin C\) is maximum if and only if b=c (you can use the area= 1/2 bc sin A to approach it).
hence for a triangle with b=c:
\(\displaystyle \sin A + \sin B \sin C= \sin A + \frac{1}{2} \cos A +\frac{1}{2}\)
if:
\(\displaystyle x=\sin A \quad then \quad f(x) =x + \frac{1}{2}\sqrt{1-x^2} + \frac {1} {2}\)
by using the derrivative of f(x), and setting f'(x)=0 we get:
\(\displaystyle x=\sin A= \frac{2}{\sqrt{5}}\)
Then the maximum value of f(x) \(\displaystyle \sin A + \sin B \sin C \quad\)) is \(\displaystyle \frac{1+\sqrt{5}}{2}\)
now we proved that for a triangle ABC \(\displaystyle \sin A + \sin B \ sin C \leq \frac{1 + \sqrt{5}}{2}\)
Let's consider a function g(x) such that:
\(\displaystyle g(x)=-x^2 + x +1\)
The roots of g(x) are : \(\displaystyle \frac{1-\sqrt{5}}{2} \quad and \quad \frac{1+\sqrt{5}}{2}\)
Hence:
\(\displaystyle for \quad every \quad x \quad: \frac{1-\sqrt{5}}{2} \leq x \leq \frac{1+\sqrt{5}}{2}\)
and since :
\(\displaystyle 0\leq \sin A + \sin B \sin C \leq \frac {1+\sqrt{5}}{2}\) (as we proved)
Then:
\(\displaystyle g(\sin A + \sin B \sin C) \geq 0\)
-->
\(\displaystyle - {(\sin A + \sin B \sin C)}^{2} + (\sin A + \sin B \sin C) +1 \geq 0\)
Thus :
\(\displaystyle (\sin A + \sin B \sin C)}^{2} \leq (\sin A + \sin B \sin C) +1\)
\(\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)
\(\displaystyle \frac{{\sin}^2 A}{{a}^2}=\frac{\sin B \sin C}{bc}\)
\(\displaystyle \sin A + \sin B \sin C= \Sin A (1 + \frac{bc}{{a}^2} \sin A)\)
For any A, and a \(\displaystyle \sin A + \sin B \sin C\) is maximum if and only if b=c (you can use the area= 1/2 bc sin A to approach it).
hence for a triangle with b=c:
\(\displaystyle \sin A + \sin B \sin C= \sin A + \frac{1}{2} \cos A +\frac{1}{2}\)
if:
\(\displaystyle x=\sin A \quad then \quad f(x) =x + \frac{1}{2}\sqrt{1-x^2} + \frac {1} {2}\)
by using the derrivative of f(x), and setting f'(x)=0 we get:
\(\displaystyle x=\sin A= \frac{2}{\sqrt{5}}\)
Then the maximum value of f(x) \(\displaystyle \sin A + \sin B \sin C \quad\)) is \(\displaystyle \frac{1+\sqrt{5}}{2}\)
now we proved that for a triangle ABC \(\displaystyle \sin A + \sin B \ sin C \leq \frac{1 + \sqrt{5}}{2}\)
Let's consider a function g(x) such that:
\(\displaystyle g(x)=-x^2 + x +1\)
The roots of g(x) are : \(\displaystyle \frac{1-\sqrt{5}}{2} \quad and \quad \frac{1+\sqrt{5}}{2}\)
Hence:
\(\displaystyle for \quad every \quad x \quad: \frac{1-\sqrt{5}}{2} \leq x \leq \frac{1+\sqrt{5}}{2}\)
and since :
\(\displaystyle 0\leq \sin A + \sin B \sin C \leq \frac {1+\sqrt{5}}{2}\) (as we proved)
Then:
\(\displaystyle g(\sin A + \sin B \sin C) \geq 0\)
-->
\(\displaystyle - {(\sin A + \sin B \sin C)}^{2} + (\sin A + \sin B \sin C) +1 \geq 0\)
Thus :
\(\displaystyle (\sin A + \sin B \sin C)}^{2} \leq (\sin A + \sin B \sin C) +1\)
mestmoha said:
Your solution here is not valid. You can't use (some) calculus (taking derivatives) in your solution.
If you can substitute some other algebra/other trigonometry to accomplish the same points,
then your solution could be checked for correctness.