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Question: How do you solve \( {\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1 \) ?...

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

Explanation

Solution

Hint : In order to determine the solution of the above trigonometric equation, rearrange the terms and use the trigonometric identity cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x . Derive the value of θ\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 2π2\pi .

Complete step-by-step answer :
We are given a trigonometric equation
sin⁡2[θ]=cos⁡2[θ]+1{\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1 and we have to find its solution
sin⁡2[θ]=cos⁡2[θ]+1{\sin ^2}\left[ \theta \right] = {\cos ^2}\left[ \theta \right] + 1
Rearranging the equation by transposing trigonometric terms on one side
cos⁡2[θ]−sin⁡2[θ]=−1{\cos ^2}\left[ \theta \right] - {\sin ^2}\left[ \theta \right] = - 1
Using the trigonometry identity cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x . SO, rewriting the above equation using this identity, we get our equation as
cos⁡2θ=−1\cos 2\theta = - 1
Taking inverse of cosine on both the sides
cos⁡−1(cos⁡2θ)=cos⁡−1θ(−1){\cos ^{ - 1}}\left( {\cos 2\theta } \right) = {\cos ^{ - 1}}\theta \left( { - 1} \right)
Since cos⁡−1(cos⁡x)=1{\cos ^{ - 1}}\left( {\cos x} \right) = 1 as they both are inverse of each other, we get
2θ=cos⁡−1θ(−1)2\theta = {\cos ^{ - 1}}\theta \left( { - 1} \right)
The value of 2θ2\theta is an angle whose cosine value is equal to -1. As we know the cos⁡π=−1→π=cos⁡−1(−1)\cos \pi = - 1 \to \pi = {\cos ^{ - 1}}\left( { - 1} \right) .
2θ=π2\theta = \pi
Generalising the above solution as the period of cosine function is 2π2\pi that means the graph of cosine function repeats itself after every 2π2\pi interval. So our general solution becomes
2θ=π+2nπ2\theta = \pi + 2n\pi where n is any integer
Dividing both sides of the equation with the number 2, we have
2θ2=12(π+2nπ) θ=π2+nπ   \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 θ=π2+nπ\theta = \dfrac{\pi }{2} + n\pi , where n is any integer.
So, the correct answer is “θ=π2+nπ\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 θ\theta
2. The domain of cosine function is in the interval [0,π]\left[ {0,\pi } \right] and the range is in the interval [−1,1]\left[ { - 1,1} \right] .
3. Even Function – A function f(x)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.
We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals, cos⁡ecθ\cos ec\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.