
Simplify $\sqrt {\left( {1 + \sin 2x} \right)} $.
Answer
493.5k+ views
Hint: Here we are given an expression and we need to simplify it. We can note that the given expression is trigonometric. The term simplify refers to make it easier or simpler. So, we need to convert our given expression into a simpler manner. To solve a trigonometric expression, we need to apply some appropriate trigonometric identities and algebraic identities.
Formula used:
The following formulas are to be used to solve the given problem.
a) ${\sin ^2}x + {\cos ^2}x = 1$
b) $\sin 2x = 2\sin x\cos x$
c)${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Complete step by step solution:
The given expression is $\sqrt {\left( {1 + \sin 2x} \right)} $.
Here, we need to substitute the known formulae in the expression.
That is in the place of $1$, we shall substitute the formula ${\sin ^2}x + {\cos ^2}x = 1$
Similarly in the place of $\sin 2x$ , we have to replace the formula $\sin 2x = 2\sin x\cos x$
Hence, we will obtain the following solution.
\[\sqrt {\left( {1 + \sin 2x} \right)} = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x + \sin 2x} \right)} \] (Here we have replaced $1$by${\sin ^2}x + {\cos ^2}x$)
\[ = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)} \] (Here we substituted $2\sin x\cos x$)
Now, we are able to note that the resultant expression inside the brackets is in the form ${a^2} + {b^2} + 2ab$.
To simplify this expression, we shall apply the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
Therefore we get\[\sqrt {\left( {1 + \sin 2x} \right)} = \sqrt {{{\left( {\sin x + \cos x} \right)}^2}} \] …..$\left( 1 \right)$ (Here $a = \sin x$ and$b = \cos x$)
Also, it is a well-known fact that the square root and the square of any number can cancel each other. For instance, if we consider $\sqrt {{{\left( 3 \right)}^2}} $ then our required answer will be $3$ .
Hence, $\left( 1 \right)$we get\[\sqrt {\left( {1 + \sin 2x} \right)} = \sin x + \cos x\] that is the required solution.
Note:
First of all, we need to check whether the given expression whether will be trigonometric or algebraic. If we are given an algebraic expression, there is no need to use trigonometric identities. But in solving a trigonometric expression, we may need to apply both algebraic and trigonometric identities.
Formula used:
The following formulas are to be used to solve the given problem.
a) ${\sin ^2}x + {\cos ^2}x = 1$
b) $\sin 2x = 2\sin x\cos x$
c)${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Complete step by step solution:
The given expression is $\sqrt {\left( {1 + \sin 2x} \right)} $.
Here, we need to substitute the known formulae in the expression.
That is in the place of $1$, we shall substitute the formula ${\sin ^2}x + {\cos ^2}x = 1$
Similarly in the place of $\sin 2x$ , we have to replace the formula $\sin 2x = 2\sin x\cos x$
Hence, we will obtain the following solution.
\[\sqrt {\left( {1 + \sin 2x} \right)} = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x + \sin 2x} \right)} \] (Here we have replaced $1$by${\sin ^2}x + {\cos ^2}x$)
\[ = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)} \] (Here we substituted $2\sin x\cos x$)
Now, we are able to note that the resultant expression inside the brackets is in the form ${a^2} + {b^2} + 2ab$.
To simplify this expression, we shall apply the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
Therefore we get\[\sqrt {\left( {1 + \sin 2x} \right)} = \sqrt {{{\left( {\sin x + \cos x} \right)}^2}} \] …..$\left( 1 \right)$ (Here $a = \sin x$ and$b = \cos x$)
Also, it is a well-known fact that the square root and the square of any number can cancel each other. For instance, if we consider $\sqrt {{{\left( 3 \right)}^2}} $ then our required answer will be $3$ .
Hence, $\left( 1 \right)$we get\[\sqrt {\left( {1 + \sin 2x} \right)} = \sin x + \cos x\] that is the required solution.
Note:
First of all, we need to check whether the given expression whether will be trigonometric or algebraic. If we are given an algebraic expression, there is no need to use trigonometric identities. But in solving a trigonometric expression, we may need to apply both algebraic and trigonometric identities.
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