Question

How do you simplify \[{\left( {\cos x + \sin x} \right)^2}\]?

Answer 1

Hint: To simplify the given expression \[{\left( {\cos x + \sin x} \right)^2}\] which is in the form of \[{\left( {a + b} \right)^2}\]. So we can use the identity given by \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] to simplify the given expression. As we know, \[{\sin ^2}x + {\cos ^2}x = 1\], using this we will further simplify it. Then using the double angle formula of trigonometry i.e., \[2\sin x\cos x = \sin 2x\] we find the result.

Complete step by step answer:
As see can that the given expression \[{\left( {\cos x + \sin x} \right)^2}\] is similar to \[{\left( {a + b} \right)^2}\] and we know that \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\].
Therefore, we can write the given expression as
\[ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = {\cos ^2}x + 2\sin x\cos x + {\sin ^2}x\]
On rewriting, we get
\[ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = {\cos ^2}x + {\sin ^2}x + 2\sin x\cos x\]
As we know from the trigonometric identities that \[{\sin ^2}x + {\cos ^2}x = 1\]. Therefore, we get
\[ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = 1 + 2\sin x\cos x\]
From the expression we can see that \[2\sin x\cos x\] belongs to the trigonometric double angle identities. So, from a double angle identity we have \[2\sin x\cos x = \sin 2x\].
So, we can replace \[2\sin x\cos x\] with \[\sin 2x\] in the above expression. So, on doing that we get
\[ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = 1 + \sin 2x\]
Therefore, the required answer or simplified form of \[{\left( {\cos x + \sin x} \right)^2}\] is \[1 + \sin 2x\].

Note:
Here, we have substituted \[a = \cos x\] and \[b = \sin x\] in \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]. This is because \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] is an identity and an identity is an equation which is always true, no matter what values are substituted whereas an equation may not be true for some values that are substituted.