Answer on Question #77063 - Math - Trigonometry
Question
Find an expression for the trigonometric equation:
F = 6 sin 8 π t − 8 cos 8 π t F = 6 \sin 8\pi t - 8 \cos 8\pi t F = 6 sin 8 π t − 8 cos 8 π t R sin ( ω t + α ) R \sin (\omega t + \alpha) R sin ( ω t + α )
Solution
Let
φ = sin − 1 ( b a 2 + b 2 ) = cos − 1 ( a a 2 + b 2 ) = tan − 1 ( b a ) \varphi = \sin^ {- 1} \left(\frac {b}{\sqrt {a ^ {2} + b ^ {2}}}\right) = \cos^ {- 1} \left(\frac {a}{\sqrt {a ^ {2} + b ^ {2}}}\right) = \tan^ {- 1} \left(\frac {b}{a}\right) φ = sin − 1 ( a 2 + b 2 b ) = cos − 1 ( a 2 + b 2 a ) = tan − 1 ( a b )
then
a ⋅ sin ( θ ) ± b ⋅ cos ( θ ) = a 2 + b 2 ( a a 2 + b 2 sin ( θ ) ± b a 2 + b 2 cos ( θ ) ) = a 2 + b 2 ( sin ( θ ) ⋅ cos ( φ ) ± cos ( θ ) ⋅ sin ( φ ) ) = a 2 + b 2 sin ( θ ± φ ) , \begin{array}{l} a \cdot \sin (\theta) \pm b \cdot \cos (\theta) = \sqrt {a ^ {2} + b ^ {2}} \left(\frac {a}{\sqrt {a ^ {2} + b ^ {2}}} \sin (\theta) \pm \frac {b}{\sqrt {a ^ {2} + b ^ {2}}} \cos (\theta)\right) \\ = \sqrt {a ^ {2} + b ^ {2}} (\sin (\theta) \cdot \cos (\varphi) \pm \cos (\theta) \cdot \sin (\varphi)) = \sqrt {a ^ {2} + b ^ {2}} \sin (\theta \pm \varphi), \\ \end{array} a ⋅ sin ( θ ) ± b ⋅ cos ( θ ) = a 2 + b 2 ( a 2 + b 2 a sin ( θ ) ± a 2 + b 2 b cos ( θ ) ) = a 2 + b 2 ( sin ( θ ) ⋅ cos ( φ ) ± cos ( θ ) ⋅ sin ( φ )) = a 2 + b 2 sin ( θ ± φ ) ,
and
F ( t ) = R ⋅ sin ( ω t + α ) = 10 ⋅ sin ( 8 π t − sin − 1 ( 0.8 ) ) , F (t) = R \cdot \sin (\omega t + \alpha) = 1 0 \cdot \sin (8 \pi t - \sin^ {- 1} (0. 8)), F ( t ) = R ⋅ sin ( ω t + α ) = 10 ⋅ sin ( 8 π t − sin − 1 ( 0.8 )) ,
where
R = 6 2 + 8 2 = 10 − a m p l i t u d e R = \sqrt {6 ^ {2} + 8 ^ {2}} = 1 0 - a m p l i t u d e R = 6 2 + 8 2 = 10 − am pl i t u d e ω = 8 ⋅ π = 2 ⋅ π ⋅ f − a n g u l a r f r e q u e n c y → f = 4 s e c − 1 − f r e q u e n c y \omega = 8 \cdot \pi = 2 \cdot \pi \cdot f - a n g u l a r f r e q u e n c y \rightarrow f = 4 s e c ^ {- 1} - f r e q u e n c y ω = 8 ⋅ π = 2 ⋅ π ⋅ f − an gu l a r f re q u e n cy → f = 4 se c − 1 − f re q u e n cy T = 1 f = 0.25 sec − p e r i o d T = \frac {1}{f} = 0. 2 5 \sec - p e r i o d T = f 1 = 0.25 sec − p er i o d α = sin − 1 ( 8 10 ) − p h a s e s h i f t → F ( 0 ) = 10 ⋅ sin ( 0 − sin − 1 ( 8 10 ) ) = − 8 \alpha = \sin^ {- 1} \left(\frac {8}{1 0}\right) - p h a s e s h i f t \rightarrow F (0) = 1 0 \cdot \sin \left(0 - \sin^ {- 1} \left(\frac {8}{1 0}\right)\right) = - 8 α = sin − 1 ( 10 8 ) − p ha ses hi f t → F ( 0 ) = 10 ⋅ sin ( 0 − sin − 1 ( 10 8 ) ) = − 8
Vertical shift is zero:
Answer: F ( t ) = R ⋅ sin ( ω t + α ) = 10 ⋅ sin ( 8 π t − sin − 1 ( 0.8 ) ) F(t) = R \cdot \sin (\omega t + \alpha) = 10 \cdot \sin (8\pi t - \sin^{-1}(0.8)) F ( t ) = R ⋅ sin ( ω t + α ) = 10 ⋅ sin ( 8 π t − sin − 1 ( 0.8 ))
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#339914 on Dec 2023