Question

How do you verify the identity${{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta $?

Answer 1

Hint: Before solving the above question let's take a brief discussion on trigonometric functions. In mathematics the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. The trigonometric functions most widely used are the sine, the cosine, and the tangent. They are used in navigation, solid mechanics, celestial, and many others.

Complete step by step solution:
The given trigonometric equation is ${{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta $. In this equation we have sin and cosine functions. In trigonometry there are many formulas related to the function. The given equation is the left and right part. We have to solve one part of the equation and if it equals to the other part then our answer is correct.
Let`s we will solve left part of the equation which is given as:
$\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1$
First we will open the bracket by using the formula${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ then we get,
$\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1$
We know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ now use this in above equation ,we get
$\begin{align}
  & \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=1+2\sin \theta \cos \theta -1 \\
 & \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=2\sin \theta \cos \theta \\
\end{align}$
Now we know that we can write $2\sin \theta \cos \theta =\sin 2\theta $ then we get,
$\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=2\sin \theta \cos \theta =\sin 2\theta $
Hence after simplify we get that left part of the equation is become equals to the right part that is ${{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta $ .

Note: We can verify our answer whether it is correct or not by solving or simplifying the right part of the equation.
The right part of the given equation is\[\sin 2\theta \], now simplifying it we get,
We know that we can write $2\sin \theta \cos \theta =\sin 2\theta $ then,
$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta $
Now add and subtract $1$ in the above equation, we get
$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta -1+1$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ now in the place of $1$ write ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta $then we get,
$\Rightarrow \sin 2\theta ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1$
Now we write ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, then we get
$\Rightarrow \sin 2\theta ={{\left( \sin \theta +\cos \theta \right)}^{2}}-1$
Hence we get the left part of the equation${{\left( \sin \theta +\cos \theta \right)}^{2}}-1$.