Question

How do you prove \[{\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - \sin x\] ?

Answer 1

Hint: We need to know the basic algebraic formulae to solve these types of questions. Also, we need to know the trigonometric formulae and conditions to make easy calculations. This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to prove the LHS side is equal to the RHS side of the given expression.

Complete step by step answer:
The given expression is shown below,
 \[{\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - \sin x \to \left( 1 \right)\]
Let’s take the LHS part to prove the given expression.
 \[LHS \Rightarrow {\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} \to \left( 2 \right)\]
We know that,
 \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
By using this formula the equation \[\left( 2 \right)\] becomes,
 \[{\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = {\cos ^2}\dfrac{x}{2} - 2\cos \dfrac{x}{2}\sin \dfrac{x}{2} + {\sin ^2}\dfrac{x}{2} \to \left( 3 \right)\]
We know that,
 \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
Here we have \[\dfrac{x}{2}\] instead of \[\theta \] . So, the equation \[\left( 3 \right)\] becomes,
 \[\left( 3 \right) \to {\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = {\cos ^2}\dfrac{x}{2} - 2\cos \dfrac{x}{2}\sin \dfrac{x}{2} + {\sin ^2}\dfrac{x}{2}\]
 \[{\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - 2\cos \dfrac{x}{2}\sin \dfrac{x}{2} \to \left( 4 \right)\]
We know that,
 \[\sin 2\theta = 2\sin \theta \cos \theta \]
Let’s take
 \[\theta = \dfrac{x}{2}\]
So, we get
 \[
  \sin 2\dfrac{x}{2} = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
  \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \to \left( 5 \right) \\
 \]
Let’s substitute the equation \[\left( 5 \right)\] in the equation \[\left( 4 \right)\] , we get
 \[\left( 4 \right) \to {\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - 2\cos \dfrac{x}{2}\sin \dfrac{x}{2}\]
 \[
  {\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - \sin x \\
  {\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = RHS \\
  LHS = RHS \\
 \]
 \[{\left( {\cos \left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right)} \right)^2} = 1 - \sin x\]
Hence proved

Note: Remember the basic trigonometric formulae and conditions to solve these types of problems. We would compare the given expression with related basic algebraic formulae and trigonometric conditions to make the easy calculation. Note that if the denominator value is zero, then the term is undefined that is the value of the term is infinity. Also, this question describes the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, note the following thing when multiply/ divide different sign terms,
1. When a negative term is multiplied/ divided with the negative term the final answer would be a positive term.
2. When a positive term is multiplied/ divided with a positive term the final answer would be a positive term.
3. When a negative term is multiplied/ divided with a positive term the final answer would be a negative term