
How do you prove ${{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x$ ?
Answer
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Hint: In the above problem, we are asked to prove the following trigonometric equation i.e. ${{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x$. For that, we are going to take the square of the L.H.S of the given equation using the algebraic identity i.e. ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. And then we will need the trigonometric identity which is equal to ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Also, we are going to require the trigonometric expression i.e. $2\sin x\cos x=\sin 2x$. Hence, using all these identities, we can prove the given equation.
Complete step by step answer:
The trigonometric equation given in the above problem which we are asked to prove is as follows:
${{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x$
Now, as you can see that L.H.S of the above equation is of the form ${{\left( a+b \right)}^{2}}$ so we can use the algebraic identity which is equal to:
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Applying the above identity in the L.H.S of the given equation we get,
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+\sin 2x$
Also, we know the trigonometric identity involving $\sin x\And \cos x$ is equal to:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Using the above property in L.H.S of the equation which we have just written above this identity we get,
$\Rightarrow 1+2\sin x\cos x=1+\sin 2x$
In the above equation, as you can see that 1 is common on both the sides so 1 will be cancelled and the above equation will look like:
$\Rightarrow 2\sin x\cos x=\sin 2x$
We also know that sine of double angle is equal to twice of sine of single angle multiplied by cosine of single angle.
$\sin 2x=2\sin x\cos x$
In the above, $x$ is the single angle and $2x$ is double of the single angle $x$. Using the above identity the equation we have written above i.e. $2\sin x\cos x=\sin 2x$ is true.
As L.H.S = R.H.S so we have proved the equation given in the above problem.
Note: You can remember the identity of sine of double angle as follows:
$\sin 2x=2\sin x\cos x$
The trick to remember the 2 in the $2\sin x\cos x$ is the 2 in $\sin 2x$. Generally, you might forget the 2 in $2\sin x\cos x$ so using this trick you cannot forget it.
Complete step by step answer:
The trigonometric equation given in the above problem which we are asked to prove is as follows:
${{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x$
Now, as you can see that L.H.S of the above equation is of the form ${{\left( a+b \right)}^{2}}$ so we can use the algebraic identity which is equal to:
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Applying the above identity in the L.H.S of the given equation we get,
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+\sin 2x$
Also, we know the trigonometric identity involving $\sin x\And \cos x$ is equal to:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Using the above property in L.H.S of the equation which we have just written above this identity we get,
$\Rightarrow 1+2\sin x\cos x=1+\sin 2x$
In the above equation, as you can see that 1 is common on both the sides so 1 will be cancelled and the above equation will look like:
$\Rightarrow 2\sin x\cos x=\sin 2x$
We also know that sine of double angle is equal to twice of sine of single angle multiplied by cosine of single angle.
$\sin 2x=2\sin x\cos x$
In the above, $x$ is the single angle and $2x$ is double of the single angle $x$. Using the above identity the equation we have written above i.e. $2\sin x\cos x=\sin 2x$ is true.
As L.H.S = R.H.S so we have proved the equation given in the above problem.
Note: You can remember the identity of sine of double angle as follows:
$\sin 2x=2\sin x\cos x$
The trick to remember the 2 in the $2\sin x\cos x$ is the 2 in $\sin 2x$. Generally, you might forget the 2 in $2\sin x\cos x$ so using this trick you cannot forget it.
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