Answer
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Hint: From the given question we have to find the equation of the sine function whose amplitude is 4 and period of pi and phase shift pi/2 to the right and vertical displacement 6 units down. The standard form of a sine function is \[\Rightarrow y=a\sin \left[ b\left( x-h \right) \right]+k\] where $a$ is the amplitude, $2\pi /b$ is the period, $h$ is the phase shift and $k$ is the vertical displacement.
Complete step-by-step answer:
Now we will observe the standard form of a sine function carefully and we will consider the given values in the question which are amplitude which is 4 and the given period which is pi and the given phase shift is pi/2 to the right and the vertical displacement of 6 units.
From comparing we get,
\[\Rightarrow a=4\]
Here \[a=4\]is the amplitude of our required sine function and we know that fron standard sine function,
\[\Rightarrow \dfrac{2\pi }{b}=period\]
\[\Rightarrow \dfrac{2\pi }{b}=\pi \]
\[\Rightarrow b=2\]
From the question the given value of the phase shift is\[\dfrac{\pi }{2}\] and by comparing the given phase shift with the standard form of sine function we can say by comparing the value of,
\[\Rightarrow h=\dfrac{\pi }{2}\]
From comparing the standard form of sine function and in the given question we can say,
\[\Rightarrow k=-6\]
Therefore the equation of the sine function from the derived values by comparing with standard sine function we can say,
\[\Rightarrow y=4\sin \left[ 2\left( x-\dfrac{\pi }{2} \right) \right]-6\]
Note: We should be very careful and must be familiar with the general standard form \[\Rightarrow y=a\sin \left[ b\left( x-h \right) \right]+k\] equations and must be very careful in calculating the values. Similarly problems of these kinds can be calculated by comparing to standard form and reduce the given equations in an easy way so that it can be compared in an accurate manner. Similarly the standard form of a cosine function is \[\Rightarrow y=a\cos \left[ b\left( x-h \right) \right]+k\] where $a$ is the amplitude, $2\pi /b$ is the period, $h$ is the phase shift and $k$ is the vertical displacement.
Complete step-by-step answer:
Now we will observe the standard form of a sine function carefully and we will consider the given values in the question which are amplitude which is 4 and the given period which is pi and the given phase shift is pi/2 to the right and the vertical displacement of 6 units.
From comparing we get,
\[\Rightarrow a=4\]
Here \[a=4\]is the amplitude of our required sine function and we know that fron standard sine function,
\[\Rightarrow \dfrac{2\pi }{b}=period\]
\[\Rightarrow \dfrac{2\pi }{b}=\pi \]
\[\Rightarrow b=2\]
From the question the given value of the phase shift is\[\dfrac{\pi }{2}\] and by comparing the given phase shift with the standard form of sine function we can say by comparing the value of,
\[\Rightarrow h=\dfrac{\pi }{2}\]
From comparing the standard form of sine function and in the given question we can say,
\[\Rightarrow k=-6\]
Therefore the equation of the sine function from the derived values by comparing with standard sine function we can say,
\[\Rightarrow y=4\sin \left[ 2\left( x-\dfrac{\pi }{2} \right) \right]-6\]
Note: We should be very careful and must be familiar with the general standard form \[\Rightarrow y=a\sin \left[ b\left( x-h \right) \right]+k\] equations and must be very careful in calculating the values. Similarly problems of these kinds can be calculated by comparing to standard form and reduce the given equations in an easy way so that it can be compared in an accurate manner. Similarly the standard form of a cosine function is \[\Rightarrow y=a\cos \left[ b\left( x-h \right) \right]+k\] where $a$ is the amplitude, $2\pi /b$ is the period, $h$ is the phase shift and $k$ is the vertical displacement.
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