Question

How do you solve $sin^{2}\theta – cos^{2}\theta = 0$ ?

Answer 1

In this question, we need to solve $sin^{2}\theta – cos^{2}\theta = 0$. The basic trigonometric functions are sine , cosine and tangent. Sine is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. Similarly, cosine is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . Here we need to find the value of $sin^{2}\theta – cos^{2}\theta = 0$ is equal to $0$ . With the help of the Trigonometric functions , we can find the value of $sin^{2}\theta – cos^{2}\theta = 0$
Formula used :
$cos^{2}\theta\ - sin^{2}\theta = cos\ 2\theta$

Complete step by step solution:
Given,
$sin^{2}\theta – cos^{2}\theta = 0$
First we can consider the left part of the given expression,
$\Rightarrow sin^{2}\theta – cos^{2}\theta$
By taking $( - )$ outside from $sin^{2}\theta – cos^{2}\theta$ , we get ,
$\Rightarrow - (cos^{2}\theta\ - \ sin^{2}\theta)$
We know that $cos^{2}\theta\ - sin^{2}\theta = cos\ 2\theta$
Thus we get,
$\Rightarrow- (cos\ 2\theta)$
Since cosine is an even function,
We get,
$\Rightarrow (\cos\ 2\theta)$
Now we need to find the values of $\theta$,
We can tell $cos\ 2\theta = 0$ if the value of $2\theta$ is $\dfrac{\pi}{2} + 2n\pi$ , $\dfrac{3\pi}{2} + 2n\pi$ etc…
We note that the period of cosine function is $\pi$ .
Thus we get $sin^{2}\theta – cos^{2}\theta = 0\$ if the value of $\theta$ is $\dfrac{\pi}{4} + n\pi$ , $\dfrac{3\pi}{4} + n\pi$ etc…
**Final answer :
$sin^{2}\theta – cos^{2}\theta = 0\$ if the value of $\theta$ is $\dfrac{\pi}{4} + n\pi$ , $\dfrac{3\pi}{4} + n\pi$ etc… **

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions.