Question

How do you solve $ {\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1 $ ?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation, rearrange the terms and use the trigonometric identity $ {\cos ^2}x - {\sin ^2}x = \cos 2x $ . Derive the value of $ \theta $ by taking inverse of cosine on both sides of the equation and obtain a generalised solution as the function cosine has a period of $ 2\pi $ .

Complete step-by-step answer:
We are given a trigonometric equation
 $ {\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1 $ and we have to find its solution
 $ {\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1 $
Rearranging the equation by transposing trigonometric terms on one side
 $ {\cos ^2}\left[ \theta \right] - {\sin ^2}\left[ \theta \right] = - 1 $
Using the trigonometry identity $ {\cos ^2}x - {\sin ^2}x = \cos 2x $ . SO, rewriting the above equation using this identity, we get our equation as
 $ \cos 2\theta = - 1 $
Taking inverse of cosine on both the sides
 $ {\cos ^{ - 1}}\left( {\cos 2\theta } \right) = {\cos ^{ - 1}}\theta \left( { - 1} \right) $
Since $ {\cos ^{ - 1}}\left( {\cos x} \right) = 1 $ as they both are inverse of each other, we get
 $ 2\theta = {\cos ^{ - 1}}\theta \left( { - 1} \right) $
The value of $ 2\theta $ is an angle whose cosine value is equal to -1. As we know the $ \cos \pi = - 1 \to \pi = {\cos ^{ - 1}}\left( { - 1} \right) $ .
 $ 2\theta = \pi $
Generalising the above solution as the period of cosine function is $ 2\pi $ that means the graph of cosine function repeats itself after every $ 2\pi $ interval. So our general solution becomes
 $ 2\theta = \pi + 2n\pi $ where n is any integer
Dividing both sides of the equation with the number 2, we have
 $
  \dfrac{{2\theta }}{2} = \dfrac{1}{2}\left( {\pi + 2n\pi } \right) \\
  \theta = \dfrac{\pi }{2} + n\pi \;
  $
Therefore, the solution of the given trigonometric equation is $ \theta = \dfrac{\pi }{2} + n\pi $ , where n is any integer.
So, the correct answer is “$ \theta = \dfrac{\pi }{2} + n\pi $ ”.

Note: 1. Generalisation of the solution is compulsory as we have not given any restrictions or interval for the value of $ \theta $
2. The domain of cosine function is in the interval $ \left[ {0,\pi } \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .
3. Even Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = f(x) $ for all x in its domain.
Odd Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = - f(x) $ for all x in its domain.
We know that $ \sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, $ \sin \theta $ and $ \tan \theta $ and their reciprocals, $ \cos ec\theta $ and $ \cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.