Can you please help me simplify \[\left( \sin\left( x \right) + cos\left( x \right) \right)^{2}\] ?

Hint: In this question, we need to find the value of \[\left( \sin\left( x \right) + cos\left( x \right) \right)^{2}\] . On looking the given expression \[\left( \sin\left( x \right) + cos\left( x \right) \right)^{2}\] is in the form of \[\left( a + b \right)^{2}\] . Thus with the help of algebraic formulae we can expand the given expression. The basic trigonometric functions are Sine , cosine and tangent . Sine function is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. Similarly, cosine function is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . With the help of the Trigonometric functions , we can find the value of \[\left( \sin\left( x \right) + cos\left( x \right) \right)^{2}\] . At first, we can expand the given expression with the help of algebraic formulas, then by using trigonometric identity and formula we can simplify the given expression.

Formula used :
1.\[sin^{2}\theta + cos^{2}\theta = 1\]
2.\[sin2\theta = 2sin\theta\ \cos\theta\]
3. \[\left( a + b \right)^{2} = a^{2} + b^{2} + 2ab\]

Complete step by step answer:
Given,
\[\Rightarrow \ \left( \sin\left( x \right) + cos\left( x \right) \right)^{2}\]
By expanding with the help of algebraic formula,
We get,
\[\left( sin\ x + \ cos\ x \right)^{2} = sin^{2}x + cos^{2}x + 2sinx\ cosx\]
From the trigonometry identity ,
\[sin^{2}\theta + cos^{2}\theta = 1\]
By substituting \[sin^{2}x + cos^{2}x = 1\]
We get,
\[\left( sin\ x + cos\ x \right)^{2} = 1 + 2sinx\ cosx\]
With the help of trigonometric double angle identities ,
\[sin2\theta = 2sin\theta\ \cos\theta\]
By substituting \[2sinx\ cosx = 2sinx\]
We get,
\[\Rightarrow \left( sin\ x + cos\ x \right)^{2} = 1 + sin2{x\ }\]
Thus we get the value of \[\left( {sin\ }\left( x \right) + \ cos\ \left( x \right) \right)^{2}\] is equal to \[1 + sin2x\]
The value of \[\left( {sin\ }\left( x \right) + \ cos\ \left( x \right) \right)^{2}\] is equal to \[1 + sin2x\]

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the algebraic formula with the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions.