
How do you find the exact value of \[\sin\dfrac{{7\pi }}{2}\]?
Answer
554.1k+ views
Hint:First try to expand the number so that it comes in a form of the trigonometric function. Then you can apply the formula and try to solve it. After that try to find out the value of sine. We can also check whether the value is positive or negative by checking that in which quadrant does the value of sine lies.
Complete step by step solution:
According to the given question, we know that \[6 + 1 = 7\]. So, if we expand \[7\], the we get:
\[\sin\dfrac{{7\pi }}{2}\]
\[ \Rightarrow \sin\dfrac{{(6 + 1)\pi }}{2}\]
By expanding the numerator, we get:
\[ \Rightarrow \sin \dfrac{{(6\pi + \pi )}}{2}\]
This expression can also be expressed as:
\[ \Rightarrow \sin \left( {\dfrac{{6\pi }}{2} + \dfrac{\pi }{2}} \right)\]
After this, we will cancel out the common terms. The terms which are common in the numerator and denominators are cancelled out. So, in this expression, \[6x\] is cancelled by \[2\] and we get
\[3x\] only at the numerator.
\[ \Rightarrow \sin \left( {3\pi + \dfrac{\pi }{2}} \right)\]
This value comes in the 3 rd Quadrant where the value of sin is negative. Now we know the
trigonometric formula that:
\[\sin (3\pi + \theta ) = - \sin \theta \]
Now, by putting the values of our expression into this formula, we get:
\[ \Rightarrow - \sin \left( {\dfrac{\pi }{2}} \right)\]
We know that the value of \[\pi = {180^ \circ }\], and when it is divided by \[2\]then we get \[{90^\circ }\]. The value of \[\sin {90^ \circ } = 1\].
So, when we simplify, we get the answer as:
\[ \Rightarrow - \sin \left( {\dfrac{\pi }{2}} \right) = - 1\]
So, the answer is \[ - 1\].
Note: The value of \[\sin \theta = 1\] when \[\theta = \dfrac{\pi }{2},\dfrac{{5\pi }}{2},\dfrac{{9\pi}}{2},\dfrac{{13\pi }}{2}.....\] and \[\sin \theta = - 1\] when \[\theta = \dfrac{{3\pi }}{2},\dfrac{{7\pi}}{2},\dfrac{{11\pi }}{2},\dfrac{{15\pi }}{2}.....\].
The values of sine, cosine and tan in the First Quadrants are positive. The values of sine are only positive in the Second Quadrant. The values of tan are only positive in the Third Quadrant and the values of cosine are only positive in the Fourth Quadrant.
Complete step by step solution:
According to the given question, we know that \[6 + 1 = 7\]. So, if we expand \[7\], the we get:
\[\sin\dfrac{{7\pi }}{2}\]
\[ \Rightarrow \sin\dfrac{{(6 + 1)\pi }}{2}\]
By expanding the numerator, we get:
\[ \Rightarrow \sin \dfrac{{(6\pi + \pi )}}{2}\]
This expression can also be expressed as:
\[ \Rightarrow \sin \left( {\dfrac{{6\pi }}{2} + \dfrac{\pi }{2}} \right)\]
After this, we will cancel out the common terms. The terms which are common in the numerator and denominators are cancelled out. So, in this expression, \[6x\] is cancelled by \[2\] and we get
\[3x\] only at the numerator.
\[ \Rightarrow \sin \left( {3\pi + \dfrac{\pi }{2}} \right)\]
This value comes in the 3 rd Quadrant where the value of sin is negative. Now we know the
trigonometric formula that:
\[\sin (3\pi + \theta ) = - \sin \theta \]
Now, by putting the values of our expression into this formula, we get:
\[ \Rightarrow - \sin \left( {\dfrac{\pi }{2}} \right)\]
We know that the value of \[\pi = {180^ \circ }\], and when it is divided by \[2\]then we get \[{90^\circ }\]. The value of \[\sin {90^ \circ } = 1\].
So, when we simplify, we get the answer as:
\[ \Rightarrow - \sin \left( {\dfrac{\pi }{2}} \right) = - 1\]
So, the answer is \[ - 1\].
Note: The value of \[\sin \theta = 1\] when \[\theta = \dfrac{\pi }{2},\dfrac{{5\pi }}{2},\dfrac{{9\pi}}{2},\dfrac{{13\pi }}{2}.....\] and \[\sin \theta = - 1\] when \[\theta = \dfrac{{3\pi }}{2},\dfrac{{7\pi}}{2},\dfrac{{11\pi }}{2},\dfrac{{15\pi }}{2}.....\].
The values of sine, cosine and tan in the First Quadrants are positive. The values of sine are only positive in the Second Quadrant. The values of tan are only positive in the Third Quadrant and the values of cosine are only positive in the Fourth Quadrant.
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