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trigonometry help
- Thread starter di95
- Start date
@di95
You already posted this here: https://www.freemathhelp.com/forum/threads/trigonometry-help.130080/#post-537597
You already posted this here: https://www.freemathhelp.com/forum/threads/trigonometry-help.130080/#post-537597
HallsofIvy
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- Jan 27, 2012
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To solve \(\displaystyle \sqrt{a}+ \sqrt{b}= c\) I would write it as \(\displaystyle \sqrt{a}= c- \sqrt{b}\) and square both sides:
\(\displaystyle a= c^2- 2c\sqrt{b}+ b\)
Now rewrite that as \(\displaystyle a- c^2- b= -2\sqrt{b}\) and square again:
\(\displaystyle (a- c^2- b)^2= 4b\).
That gets rid of the square roots.
As an alternative to Hall’s approach, we could also isolate the variable and then square
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - \sqrt{(4 sin(0.5) - 6)^2} = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - (4 sin(0.5) - 6) = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} = - 8.5 + 4sin(0.5).[/MATH]
From that point, it is obvious that there is no solution.
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - \sqrt{(4 sin(0.5) - 6)^2} = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - (4 sin(0.5) - 6) = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} = - 8.5 + 4sin(0.5).[/MATH]
From that point, it is obvious that there is no solution.
A couple of observations:
and you have dropped a 'c':
\(\displaystyle a- c^2- b= -2\boldsymbol{c}\sqrt{b}\)
[MATH]\sqrt{sin^2(0.5x) + \boldsymbol{2}sin(0.5x) + 1} - \sqrt{(4 sin(0.5) - 6)^2} = -2.5[/MATH]and [MATH]\sqrt{(4 sin(0.5) - 6)^2}[/MATH] is not (4 sin(0.5) - 6), rather (6-4 sin(0.5))
also affecting the conclusion:
"there is no solution."
There is more background, with the OP's work, in the duplicate thread mentioned above:
Surely this question is of the form \(\displaystyle \sqrt{a}- \sqrt{b}= c\)To solve \(\displaystyle \sqrt{a}+ \sqrt{b}= c\) I would write it as \(\displaystyle \sqrt{a}= c- \sqrt{b}\) and square both sides:
\(\displaystyle a= c^2- 2c\sqrt{b}+ b\)
Now rewrite that as \(\displaystyle a- c^2- b= -2\sqrt{b}\) and square again:
\(\displaystyle (a- c^2- b)^2= 4b\).
That gets rid of the square roots.
and you have dropped a 'c':
\(\displaystyle a- c^2- b= -2\boldsymbol{c}\sqrt{b}\)
A typo - the question was:As an alternative to Hall’s approach, we could also isolate the variable and then square
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - \sqrt{(4 sin(0.5) - 6)^2} = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} - (4 sin(0.5) - 6) = -2.5 \implies[/MATH]
[MATH]\sqrt{sin^2(0.5x) + sin(0.5x) + 1} = - 8.5 + 4sin(0.5).[/MATH]
From that point, it is obvious that there is no solution.
[MATH]\sqrt{sin^2(0.5x) + \boldsymbol{2}sin(0.5x) + 1} - \sqrt{(4 sin(0.5) - 6)^2} = -2.5[/MATH]and [MATH]\sqrt{(4 sin(0.5) - 6)^2}[/MATH] is not (4 sin(0.5) - 6), rather (6-4 sin(0.5))
also affecting the conclusion:
"there is no solution."
There is more background, with the OP's work, in the duplicate thread mentioned above:
Last edited:
Correction.By trial and error, [MATH]x \approx 4\pi n + 1.2431025[/MATH]
where [MATH]n[/MATH] is any integer
There is also a second solution, [MATH]x \approx 2\pi n - 1.2431025[/MATH]
where [MATH]n[/MATH] is any integer, [MATH]n \neq 0[/MATH]
There is also a second solution, [MATH]x \approx 2\pi n - 1.2431025[/MATH]where [MATH]n[/MATH] is any Odd integer, [MATH]n \neq 0[/MATH]
Yes, of course. The square root must be non-negative and so must equal[MATH]\sqrt{(4 sin(0.5) - 6)^2}[/MATH] is not (4 sin(0.5) - 6), rather (6-4 sin(0.5))
[MATH]6 - 4sin(0.5).[/MATH]