Question

How do you find the exact value of ${\sin ^2}x - {\cos ^2}x = 0$ in the interval ${0^ \circ } \leqslant x < {360^ \circ }$ ?

Answer 1

Hint: The given question involves solving of a trigonometric equation and finding value of angle x that satisfy the given equation and lie in the range of ${0^ \circ } \leqslant x < {360^ \circ }$. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step by step solution:
In the given problem, we have to solve the trigonometric equation ${\sin ^2}x - {\cos ^2}x = 0$ and find the values of x that satisfy the given equation and lie in the range of ${0^ \circ } \leqslant x < {360^ \circ }$.
So, In order to solve the given trigonometric equation${\sin ^2}x - {\cos ^2}x = 0$ , we should first take all the terms to the left side of the equation.
In the given trigonometric equation, all the terms are already on the left side of the equation. Hence, we have,
$ \Rightarrow {\sin ^2}x - {\cos ^2}x = 0$
Now, we multiply both the sides of the equation by $\left( { - 1} \right)$ in order to resemble the expression with the double angle formula of cosine, we get,
$ \Rightarrow {\cos ^2}x - {\sin ^2}x = 0$
Now, we know the double angle formula for cosine is $\cos \left( {2x} \right) = {\cos ^2}\left( x \right) - {\sin ^2}x$ . Hence, substituting ${\cos ^2}\left( x \right) - {\sin ^2}x$ as $\cos \left( {2x} \right)$, we get,
$ \Rightarrow \cos \left( {2x} \right) = 0$
The above equation represents a simple form of the trigonometric equation. We know that cosine is equal to zero at odd multiples of ${90^ \circ }$ .
So, \[2x = \left( {2n + 1} \right)\left( {{{90}^ \circ }} \right)\] where n is any integer
\[ \Rightarrow x = \dfrac{{\left( {2n + 1} \right)}}{2}\left( {{{90}^ \circ }} \right)\]
Now, we have found all the values of x that satisfy the given trigonometric equation. Now, we just have to select the values that lie in the interval ${0^ \circ } \leqslant x < {360^ \circ }$.
So, for $n = 0$ in \[x = \left( {\dfrac{{2n + 1}}{2}} \right)\left( {{{90}^ \circ }} \right)\] , we get $x = {45^ \circ }$.
For $n = 1$ in \[x = \left( {\dfrac{{2n + 1}}{2}} \right)\left( {{{90}^ \circ }} \right)\] , we get $x = {135^ \circ }$ .
For $n = 2$ in \[x = \left( {\dfrac{{2n + 1}}{2}} \right)\left( {{{90}^ \circ }} \right)\], we get $x = {225^ \circ }$ .
For $n = 3$ in \[x = \left( {\dfrac{{2n + 1}}{2}} \right)\left( {{{90}^ \circ }} \right)\], we get $x = {315^ \circ }$.
Hence, the values of x that satisfy the given trigonometric equation ${\sin ^2}x - {\cos ^2}x = 0$ and lie between the interval ${0^ \circ } \leqslant x < {360^ \circ }$ are: $x = {45^ \circ }$, ${135^ \circ }$,${225^ \circ }$ and ${315^ \circ }$ .

Note: The given trigonometric equation can also be solved by first finding the value of ${\cos ^2}\left( x \right)$ in ${\sin ^2}x - {\cos ^2}x = 0$ using ${\cos ^2}x + {\sin ^2}x = 1$and then finding the value of $\cos \left( x \right)$as $\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right)$. Then, we solve the two equations $\cos \left( x \right) = \dfrac{1}{{\sqrt 2 }}$ and $\cos \left( x \right) = - \dfrac{1}{{\sqrt 2 }}$ and find the values of x that satisfy either of the equations and lie between the interval ${0^ \circ } \leqslant x < {360^ \circ }$.