
How do you evaluate \[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right)?\]
Answer
558.9k+ views
Hint:We need to know the trigonometric table values and basic definitions \[\tan \theta \] . We need to know the value of \[\cos \left( {\dfrac{{ - 7\pi }}{4}} \right)\] and \[\sin \left( {\dfrac{{ - 7\pi }}{4}} \right)\] .
This question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the degree value of \[\pi \] terms. By having the value of \[\sin \theta \] and \[\cos \theta \] we can easily find out the value of \[\tan \theta \] .
Complete step by step solution:
The given question is shown below
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = ?\] \[ \to \left( 1 \right)\]
To solve the above equation we need to know the basic definition \[\tan \theta \] .
The above figure is used to define the \[\tan \theta \] according to the position of \[\theta
\] .
So, we get
\[\tan \theta =
\dfrac{{opposite}}{{adjacant}} \to \left( 2
\right)\]
Also, we get the definition for \[\sin \theta
\] and \[\cos \theta \] from the figure mentioned above.
We get,
\[\sin \theta = \dfrac{{opposite}}{{hypotenuse}}\] \[ \to \left( 3 \right)\]
And,
\[\cos \theta = \dfrac{{adjacant}}{{hypotenuse}}\] \[ \to \left( 4 \right)\]
Let’s divide the equation \[\left( 3 \right)\] by the equation \[\left( 4 \right)\] we get,
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\left( {\dfrac{{opposite}}{{hypotenuse}}}
\right)}}{{\left( {\dfrac{{adjacant}}{{hypotenuse}}} \right)}}\]
The above equation can also be written as,
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{hypotenuse}} \times
\dfrac{{hypotenuse}}{{adjacant}}\]
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{adjacant}} \to \left( 5 \right)\]
By comparing the equation \[\left( 5 \right)\] and \[\left( 2 \right)\] , we get
\[
\left( 2 \right) \to \tan \theta = \dfrac{{opposite}}{{adjacant}} \\
\left( 5 \right) \to \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{adjacant}} \\
\]
So,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \to \left( 6 \right)\]
We know that the value \[\theta \] is \[\left( {\dfrac{{ - 7\pi }}{4}} \right)\] (Given in the question).
From the trigonometric table value,
\[
\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
\]
So, \[\sin \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{4}} \right)\]
From the above equations, we get,
\[\sin \left( {\dfrac{{ - 7\pi }}{4}} \right) = \cos \left( {\dfrac{{ - 7\pi }}{4}} \right)\] \[ \to \left( 7
\right)\]
Let’s substitute the above equation in the equation \[\left( 6 \right)\] , we get
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = \dfrac{{\left( {\sin \dfrac{{ - 7\pi }}{4}} \right)}}{{\left(
{\cos \dfrac{{ - 7\pi }}{4}} \right)}}\]
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = 1\]
So, the final answer is,
The value of \[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right)\] is \[1\] .
Note: In these types of questions we would remember the trigonometric table values and basic definitions of \[\sin \theta ,\cos \theta \] and \[\tan \theta \] .Note that when \[\dfrac{\pi }{4}\] is involved in \[\theta \] value, the value of \[\tan \theta \] is always \[1\] . When we have the fraction term in the denominator, the fraction term of the denominator will come to the position of the numerator. If we have a fraction term in the numerator, the denominator of the fraction term will come to the position of the denominator as follows,
\[\dfrac{{\left( {\dfrac{x}{y}} \right)}}{{\left( {\dfrac{z}{l}} \right)}} = \left( {\dfrac{x}{y} \times
\dfrac{l}{z}} \right)\]
This question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the degree value of \[\pi \] terms. By having the value of \[\sin \theta \] and \[\cos \theta \] we can easily find out the value of \[\tan \theta \] .
Complete step by step solution:
The given question is shown below
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = ?\] \[ \to \left( 1 \right)\]
To solve the above equation we need to know the basic definition \[\tan \theta \] .
The above figure is used to define the \[\tan \theta \] according to the position of \[\theta
\] .
So, we get
\[\tan \theta =
\dfrac{{opposite}}{{adjacant}} \to \left( 2
\right)\]
Also, we get the definition for \[\sin \theta
\] and \[\cos \theta \] from the figure mentioned above.
We get,
\[\sin \theta = \dfrac{{opposite}}{{hypotenuse}}\] \[ \to \left( 3 \right)\]
And,
\[\cos \theta = \dfrac{{adjacant}}{{hypotenuse}}\] \[ \to \left( 4 \right)\]
Let’s divide the equation \[\left( 3 \right)\] by the equation \[\left( 4 \right)\] we get,
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\left( {\dfrac{{opposite}}{{hypotenuse}}}
\right)}}{{\left( {\dfrac{{adjacant}}{{hypotenuse}}} \right)}}\]
The above equation can also be written as,
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{hypotenuse}} \times
\dfrac{{hypotenuse}}{{adjacant}}\]
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{adjacant}} \to \left( 5 \right)\]
By comparing the equation \[\left( 5 \right)\] and \[\left( 2 \right)\] , we get
\[
\left( 2 \right) \to \tan \theta = \dfrac{{opposite}}{{adjacant}} \\
\left( 5 \right) \to \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{opposite}}{{adjacant}} \\
\]
So,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \to \left( 6 \right)\]
We know that the value \[\theta \] is \[\left( {\dfrac{{ - 7\pi }}{4}} \right)\] (Given in the question).
From the trigonometric table value,
\[
\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
\]
So, \[\sin \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{4}} \right)\]
From the above equations, we get,
\[\sin \left( {\dfrac{{ - 7\pi }}{4}} \right) = \cos \left( {\dfrac{{ - 7\pi }}{4}} \right)\] \[ \to \left( 7
\right)\]
Let’s substitute the above equation in the equation \[\left( 6 \right)\] , we get
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = \dfrac{{\left( {\sin \dfrac{{ - 7\pi }}{4}} \right)}}{{\left(
{\cos \dfrac{{ - 7\pi }}{4}} \right)}}\]
\[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right) = 1\]
So, the final answer is,
The value of \[\tan \left( {\dfrac{{ - 7\pi }}{4}} \right)\] is \[1\] .
Note: In these types of questions we would remember the trigonometric table values and basic definitions of \[\sin \theta ,\cos \theta \] and \[\tan \theta \] .Note that when \[\dfrac{\pi }{4}\] is involved in \[\theta \] value, the value of \[\tan \theta \] is always \[1\] . When we have the fraction term in the denominator, the fraction term of the denominator will come to the position of the numerator. If we have a fraction term in the numerator, the denominator of the fraction term will come to the position of the denominator as follows,
\[\dfrac{{\left( {\dfrac{x}{y}} \right)}}{{\left( {\dfrac{z}{l}} \right)}} = \left( {\dfrac{x}{y} \times
\dfrac{l}{z}} \right)\]
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