Question

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

Answer 1

Hint: In such types of trigonometry questions we have to use the Pythagoras theorem and the basic definitions of sine and cos.
In this regards we have to use the following trigonometric formula:
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$

Complete step by step answer:
We know that according to Pythagoras theorem
$\sin \theta =\dfrac{o}{h}$
And
$\cos \theta =\dfrac{a}{h}$
Where
$o=$ Side of the opposite angle
$a=$ Side adjacent to the angle
$h=$ Hypognathous of the right angle triangle
We know that the Pythagoras theorem states that
${{h}^{2}}={{a}^{2}}+{{o}^{2}}$
After rearranging the terms of the above expression we get
$1=\dfrac{{{a}^{2}}}{{{h}^{2}}}+\dfrac{{{o}^{2}}}{{{h}^{2}}}$
Now from the definitions of $\sin \theta $ and $\cos \theta $we get
$1={{\sin }^{2}}\theta +{{\cos }^{2}}\theta $
After rearranging the terms of the above expression we get
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \ \cdots \cdots \left( 1 \right)$
We know that from the algebraic formula
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Therefore,
$1-{{\cos }^{2}}\theta =\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)$
Now the equation (1) can be written in the following form
${{\sin }^{2}}\theta =\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)$
Hence, the given expression is proved.

Note:
In such types of questions, we have seen that we can easily solve these types of questions with the help of trigonometric formulas. So, it is important for a student to remember formulas and important trigonometric properties. There is also an alternative method to solve this question.
We know that
$\left( 1+\cos \theta \right)\left( 1+\cos \theta \right)=1-{{\cos }^{2}}\theta \ \cdots \cdots \left( 2 \right)$
We also know that
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\ $
Further rearranging the terms of the above expression we get
$1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \ \cdots \cdots \left( 3 \right)$
Now substitute the necessary values from equation (3) to equation (2) we get
$\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)={{\sin }^{2}}\theta $
This is the proof of the given expression by the alternative method.