How do you simplify \[\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)\]?
Answer
625.5k+ views
Hint: we will use basic inverse trigonometric functions to solve the problem. We will assign the inverse sin value to any variable and then we will find its value. After that we have to substitute the value in the place of inverse sin and then we will simplify it further to arrive at the solution.
Complete step-by-step solution:
Given equation is
\[\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)\]
Now we have to assign inverse sin value to any variable
Let us \[\theta \] as the variable. So we will take
\[\Rightarrow \theta =\arcsin \left( \dfrac{1}{3} \right)......\left( 1 \right)\]
From this we can write
\[\Rightarrow \sin \theta =\left( \dfrac{1}{3} \right)\]
Using \[\sin \theta \] we can find the value of \[\cos \theta \].
We have the basic trigonometry identity
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
From this we can write
\[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]
Using this we can find the \[\cos \theta \].
We will get
\[\Rightarrow \cos \theta =\sqrt{1-{{\left( \dfrac{1}{3} \right)}^{2}}}\]
By simplifying it, we will get
\[\Rightarrow \cos \theta =\sqrt{1-\dfrac{1}{9}}\]
\[\Rightarrow \cos \theta =\sqrt{\dfrac{8}{9}}\]
We can write \[\sqrt{\dfrac{8}{9}}\] as \[\dfrac{2\sqrt{2}}{3}\] because square root of 9 is 3 and we can write 8 as \[4\times 2\].
After simplifying it we will get
\[\Rightarrow \cos \theta =\dfrac{2\sqrt{2}}{3}\]
Now in equation 1 we have the value of \[\theta \]. So we have to substitute that \[\theta \] value here to get the answer.
We have \[\theta \] as
\[\theta =\arcsin \left( \dfrac{1}{3} \right)\]
Let us substitute this in the \[\cos \theta \].
After substituting we will get
\[\Rightarrow \cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)\]
We have the value of \[\cos \theta \] as \[\dfrac{2\sqrt{2}}{3}\].
So we get the value as
\[\Rightarrow \cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{2\sqrt{2}}{3}\]
So by simplifying the given question we got \[\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{2\sqrt{2}}{3}\].
Note: we can solve this in other methods by drawing triangles in the plane with vertices and the origin. After drawing using the angles we got we will simplify the expression further to get the solution. But the solution described above is the simplest way of solving.
Complete step-by-step solution:
Given equation is
\[\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)\]
Now we have to assign inverse sin value to any variable
Let us \[\theta \] as the variable. So we will take
\[\Rightarrow \theta =\arcsin \left( \dfrac{1}{3} \right)......\left( 1 \right)\]
From this we can write
\[\Rightarrow \sin \theta =\left( \dfrac{1}{3} \right)\]
Using \[\sin \theta \] we can find the value of \[\cos \theta \].
We have the basic trigonometry identity
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
From this we can write
\[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]
Using this we can find the \[\cos \theta \].
We will get
\[\Rightarrow \cos \theta =\sqrt{1-{{\left( \dfrac{1}{3} \right)}^{2}}}\]
By simplifying it, we will get
\[\Rightarrow \cos \theta =\sqrt{1-\dfrac{1}{9}}\]
\[\Rightarrow \cos \theta =\sqrt{\dfrac{8}{9}}\]
We can write \[\sqrt{\dfrac{8}{9}}\] as \[\dfrac{2\sqrt{2}}{3}\] because square root of 9 is 3 and we can write 8 as \[4\times 2\].
After simplifying it we will get
\[\Rightarrow \cos \theta =\dfrac{2\sqrt{2}}{3}\]
Now in equation 1 we have the value of \[\theta \]. So we have to substitute that \[\theta \] value here to get the answer.
We have \[\theta \] as
\[\theta =\arcsin \left( \dfrac{1}{3} \right)\]
Let us substitute this in the \[\cos \theta \].
After substituting we will get
\[\Rightarrow \cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)\]
We have the value of \[\cos \theta \] as \[\dfrac{2\sqrt{2}}{3}\].
So we get the value as
\[\Rightarrow \cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{2\sqrt{2}}{3}\]
So by simplifying the given question we got \[\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{2\sqrt{2}}{3}\].
Note: we can solve this in other methods by drawing triangles in the plane with vertices and the origin. After drawing using the angles we got we will simplify the expression further to get the solution. But the solution described above is the simplest way of solving.
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