Question

How does one solve ${\cos ^2}\theta = 1$ ?

Answer 1

Hint: For solving this particular question , you have to simplify the given expression by reordering the equation , applying trigonometric identity , taking square root and then use the concept of periodicity of a sine function.

Complete solution step by step:
It is given that ${\cos ^2}\theta = 1$ ,
And we have to solve for $\theta $ .
Now taking ,
${\cos ^2}\theta = 1$
We can write this as ,
$ \Rightarrow {\cos ^2}\theta - 1 = 0$
Now do reordering as follow ,
$ \Rightarrow - 1 + {\cos ^2}\theta = 0$
Now take negative signs common from the left hand side . we will get ,
$ \Rightarrow - (1 - {\cos ^2}\theta ) = 0$
Now applying the trigonometric identity that is $1 - {\cos ^2}\theta = {\sin ^2}\theta $ , we will get as follow,
$ \Rightarrow - {\sin ^2}\theta = 0$
Or we can write this as ,
$ \Rightarrow {\sin ^2}\theta = 0$
Now taking square root both the side ,
$ \Rightarrow \sin \theta = \pm 0$
Or
$ \Rightarrow \sin \theta = 0$
We know that $\sin \theta $ could be a periodic performance that oscillates over a regular interval. It crosses the coordinate axis (i.e.., $\sin \theta $ is zero) at$x = 0,\pi ,2\pi $ within the domain $[0,2\pi ]$ , and continues to cross the coordinate axis at each whole number multiple of $\pi $.
Therefore, $\theta = \pi n$ where $n$ is any integer number .

Additional Information: In arithmetic, pure mathematics identities are equalities that involve pure mathematics functions and are true for every worth of the occurring variables that every aspect of the equality are outlined. Geometrically, these are identities involving sure functions of one or additional angles.

Note: Identities are helpful whenever expressions involving pure mathematics functions should be simplified. An important application is that the combination of non-trigonometric functions: a typical technique involves initial mistreatment the substitution rule with a mathematical relation, then simplifying the ensuing integral with a pure mathematics identity.