Question

How do you solve ${\sin ^2}A = \cos A - 1$ ?

Answer 1

Hint : In order to determine the solution of the above trigonometric equation replace the ${\sin ^2}x$ using the identity of trigonometry ${\sin ^2}x = 1 - {\cos ^2}x$ . We will obtain a quadratic equation. Factorise the quadratic equation using splitting by the middle term and then equate every factor equal to zero to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation ${\sin ^2}A = \cos A - 1$ and we have to find the solution
${\sin ^2}A = \cos A - 1$
Using the trigonometry ${\sin ^2}x = 1 - {\cos ^2}x$ to replace ${\sin ^2}x$ from the equation ,we get
$1 - {\cos ^2}A = \cos A - 1$
Rearranging the terms in the standard quadratic form $a{x^2} + bx + c$
${\cos ^2}A + \cos A - 2 = 0$
Factoring the above quadratic equation by using the middle term splitting method by splitting $\cos A\,as\,2\cos A - \cos A$ . Our equation now becomes
${\cos ^2}A + 2\cos A - \cos A - 2 = 0$
Pulling out common from first two terms and last two terms
$\cos A\left( {\cos A + 2} \right) - 1\left( {\cos A + 2} \right) = 0$
Finding the common binomial parenthesis, the equation becomes
$\left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0$ ----(1)
Dividing both sides of equation with $\left( {\cos A - 1} \right)$ , we have

$$\dfrac{1}{{\left( {\cos A - 1} \right)}} \times \left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0 \times \dfrac{1}{{\left( {\cos A - 1} \right)}} \\\ \left( {\cos A + 2} \right) = 0 \\\ \cos A = - 2 \;$$

Since the range of cosine function is in the interval $\left[ { - 1,1} \right]$ so the above is not possible
Now dividing the equation (1) with $\left( {\cos A + 2} \right)$ , we get

$$\dfrac{1}{{\left( {\cos A + 2} \right)}} \times \left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0 \times \dfrac{1}{{\left( {\cos A + 2} \right)}} \\\ \left( {\cos A - 1} \right) = 0 \\\ \Rightarrow \cos A = 1 \\\ A = {\cos ^{ - 1}}\left( 1 \right) \;$$

The value of A is an angle having cosine value 1. Since we know $\cos 0 = 1 \to 0 = {\cos ^{ - 1}}\left( 1 \right)$ .
And we know the period of cosine function is $2\pi$ as it repeats itself after every $2\pi$ interval. We have value of A as
$A = 2n\pi$ where n is any integer value
Therefore the solution of the given trigonometric equation is $A = 2n\pi$ , n is any integer.
So, the correct answer is “ $A = 2n\pi$ ”.

Note : Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$ .If $a = 0$ then the equation will become linear equation and will no more quadratic .
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of cosine function is $2\pi$ .
3. The domain of cosine function is in the interval $\left[ {0,\pi } \right]$ and the range is in the interval $\left[ { - 1,1} \right]$ .
2.Don’t forget to rearrange quadratic equations in the standard form.
3. Write the factors when the middle term is split using the proper sign.