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Sagot :
Answer:
Approximately [tex]9.62[/tex].
Explanation:
[tex]y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0][/tex].
[tex]y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)][/tex].
Notice that sine waves [tex]y_1[/tex] and [tex]y_2[/tex] share the same frequency and wavelength. The only distinction between these two waves is the [tex](-0.250)[/tex] in [tex]y_2\![/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would still be a sine wave. The amplitude of [tex](y_1 + y_2)\![/tex] could be found without using calculus.
Consider the sum-of-angle identity for sine:
[tex]\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)[/tex].
Compare the expression [tex]\sin(a + b)[/tex] to [tex]y_2[/tex]. Let [tex]a = (4.35\, x - 1270)[/tex] and [tex]b = (-0.250)[/tex]. Apply the sum-of-angle identity of sine to rewrite [tex]y_2\![/tex].
[tex]\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would become:
[tex]\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Consider: would it be possible to find [tex]m[/tex] and [tex]c[/tex] that satisfy the following hypothetical equation?
[tex]\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Simplify this hypothetical equation:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}[/tex].
Apply the sum-of-angle identity of sine to rewrite the left-hand side:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}[/tex].
Compare this expression with the right-hand side. For this hypothetical equation to hold for all real [tex]x[/tex] and [tex]t[/tex], the following should be satisfied:
[tex]\displaystyle 1 + \cos(-0.250) = m\, \cos(c)[/tex], and
[tex]\displaystyle \sin(-0.250) = m\, \sin(c)[/tex].
Consider the Pythagorean identity. For any real number [tex]a[/tex]:
[tex]{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2[/tex].
Make use of the Pythagorean identity to solve this system of equations for [tex]m[/tex]. Square both sides of both equations:
[tex]\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2[/tex].
[tex]\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2[/tex].
Take the sum of these two equations.
Left-hand side:
[tex]\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}[/tex].
Right-hand side:
[tex]\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}[/tex].
Therefore:
[tex]m^2 = 2 + 2\, \cos(-0.250)[/tex].
[tex]m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98[/tex].
Substitute [tex]m = \sqrt{2 + 2\, \cos(-0.250)}[/tex] back to the system to find [tex]c[/tex]. However, notice that the exact value of [tex]c\![/tex] isn't required for finding the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex].
(Side note: one possible value of [tex]c[/tex] is [tex]\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125[/tex] radians.)
As long as [tex]\! c[/tex] is a real number, the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex] would be equal to the absolute value of [tex](4.85\, m)[/tex].
Therefore, the amplitude of [tex](y_1 + y_2)[/tex] would be:
[tex]\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}[/tex].
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