
How do you simplify $\sin \left( t+\dfrac{\pi }{2} \right)$?
Answer
559.2k+ views
Hint: We will see the definition of a sine function. We will look at the formula for the sine function of the sum of two angles. We will use this formula to expand the given function. Then we will find the value of the trigonometric function of the standard angle needed to substitute in the formula. We will simplify the obtained expression and get the required value.
Complete step-by-step solution:
The sine function is defined as $\sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}}$ for an acute angle $\theta $, in a right angled triangle.
We have formulae for the expansion of trigonometric functions of sum of two angles or difference of two angles. The formula for the sine function of the sum of two angles is given as,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Let us substitute $A=t$ and $B=\dfrac{\pi }{2}$ in the above formula. We get the expansion of the given function in the following manner,
$\sin \left( t+\dfrac{\pi }{2} \right)=\sin t\cos \dfrac{\pi }{2}+\cos t\sin \dfrac{\pi }{2}$
We know that the angle $\dfrac{\pi }{2}$ is a standard angle. The value of the sine function and cosine function for this angle is given as $\sin \dfrac{\pi }{2}=1$ and $\cos \dfrac{\pi }{2}=0$. Substituting these values in the above equation, we get the following,
$\begin{align}
& \sin \left( t+\dfrac{\pi }{2} \right)=\sin t\left( 0 \right)+\cos t\left( 1 \right) \\
& \therefore \sin \left( t+\dfrac{\pi }{2} \right)=\cos t \\
\end{align}$
Therefore, the simplified form of the given equation is $\sin \left( t+\dfrac{\pi }{2} \right)=\cos t$.
Note: The simplification that we have obtained is an identity and we have seen the proof for this identity. The sine function can also be defined using the unit circle in the cartesian plane. The x-coordinate and the y-coordinate of the point on the circle represent the cosine function value and the sine function value respectively.
Complete step-by-step solution:
The sine function is defined as $\sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}}$ for an acute angle $\theta $, in a right angled triangle.
We have formulae for the expansion of trigonometric functions of sum of two angles or difference of two angles. The formula for the sine function of the sum of two angles is given as,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Let us substitute $A=t$ and $B=\dfrac{\pi }{2}$ in the above formula. We get the expansion of the given function in the following manner,
$\sin \left( t+\dfrac{\pi }{2} \right)=\sin t\cos \dfrac{\pi }{2}+\cos t\sin \dfrac{\pi }{2}$
We know that the angle $\dfrac{\pi }{2}$ is a standard angle. The value of the sine function and cosine function for this angle is given as $\sin \dfrac{\pi }{2}=1$ and $\cos \dfrac{\pi }{2}=0$. Substituting these values in the above equation, we get the following,
$\begin{align}
& \sin \left( t+\dfrac{\pi }{2} \right)=\sin t\left( 0 \right)+\cos t\left( 1 \right) \\
& \therefore \sin \left( t+\dfrac{\pi }{2} \right)=\cos t \\
\end{align}$
Therefore, the simplified form of the given equation is $\sin \left( t+\dfrac{\pi }{2} \right)=\cos t$.
Note: The simplification that we have obtained is an identity and we have seen the proof for this identity. The sine function can also be defined using the unit circle in the cartesian plane. The x-coordinate and the y-coordinate of the point on the circle represent the cosine function value and the sine function value respectively.
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