Question

Solve the following equation for A and B.
$\cos \left( A-B \right)-\cos \left( 180-\left( A+B \right) \right)=0$

Answer 1

Hint: At first check $\left( 180-\left( A+B \right) \right)$ lies in which coordinate. According to that change the sign of the cos function if needed. Then simplify the equation by using the following formulas:
$\cos \left( A+B \right)=\operatorname{cosA}\operatorname{cosB}-\operatorname{sinA}\operatorname{sinB},cos\left( A-B \right)=cosAcosB+sinAsinB$

Complete step-by-step answer:

The given equation is:
$\cos \left( A-B \right)-\cos \left( 180-\left( A+B \right) \right)=0$
$\Rightarrow \cos \left( A-B \right)-\cos \left( \pi -\left( A+B \right) \right)=0.......(1)$
The angle $\left( \pi -\left( A+B \right) \right)$ lies in the second coordinate. We know that, in second coordinate, cos is negative. When we have $\left( \pi \pm \theta \right)$ as an angle, cos will remain the same. Therefore, $\cos \left( \pi -\left( A+B \right) \right)=-\cos \left( A+B \right)$
If we put this value in equation (1) we will get,
$\Rightarrow \cos \left( A-B \right)-\left[ -\cos \left( A+B \right) \right]=0$
$\Rightarrow \cos \left( A-B \right)+\cos \left( A+B \right)=0.....(2)$
Now we know that,
$\cos \left( A+B \right)=\operatorname{cosA}\operatorname{cosB}-\operatorname{sinA}\sin B,cos\left( A-B \right)=cosAcosB+sinsAinB$
We can put these values in equation (2). Therefore,
$\Rightarrow \left( \operatorname{cosA}\operatorname{cosB}+\operatorname{sinA}\operatorname{sinB} \right)+\left( \operatorname{cosA}\operatorname{cosB}-\operatorname{sinA}\sin B \right)=0$
Now, we can cancel out the opposite terms and add the similar terms.
$\Rightarrow 2\operatorname{cosA}\cos B=0$
$\Rightarrow \operatorname{cosA}\cos B=0$
Therefore either $\cos A=0$ or $\cos B=0$.
Let us first take, $\cos A=0.......(3)$
We know that if we multiply $\dfrac{\pi }{2}$ by any odd number, the cos value of that angle always will be zero. That means, $\cos \dfrac{\pi }{2}=0,\cos \dfrac{3\pi }{2}=0,\cos \dfrac{5\pi }{2}=0,...$
Therefore, $\cos \dfrac{\left( 2n-1 \right)\pi }{2}=0$, where n is any natural number.
Now we can put $\cos \dfrac{\left( 2n-1 \right)\pi }{2}$ in the left hand side of equation (3).
$\Rightarrow \cos A=\cos \dfrac{\left( 2n-1 \right)\pi }{2},n=1,2,3,...$
Multiply both side of the equation by ${{\cos }^{-1}}$,
$\Rightarrow A={{\cos }^{-1}}\left( \cos \left( \dfrac{\left( 2n-1 \right)\pi }{2} \right) \right)$
$\Rightarrow A=\dfrac{\left( 2n-1 \right)\pi }{2}$, where n is any natural number.
Either we can take, $\operatorname{cosB}=0$
Similarly we can write,
$\Rightarrow \operatorname{cosB}=\cos \dfrac{\left( 2n-1 \right)\pi }{2},n=1,2,3,...$
Multiply both side of the equation by ${{\cos }^{-1}}$,
$\Rightarrow B={{\cos }^{-1}}\left( \cos \left( \dfrac{\left( 2n-1 \right)\pi }{2} \right) \right)$
$\Rightarrow B=\dfrac{\left( 2n-1 \right)\pi }{2}$, where n is any natural number.
Therefore, the solution is $A=\dfrac{\left( 2n-1 \right)\pi }{2},B=\dfrac{\left( 2n-1 \right)\pi }{2}$.
Note: We generally make mistakes while we change the sign of a trigonometric function for a particular coordinate. We need to remember that all the trigonometric functions are always positive in the first coordinate. In the second coordinate only sin and cosec functions are positive. In the third coordinate only tan and cot are positive. In the fourth coordinate only cos and sec are positive.
For the angles in the form of $\left( n\pi +\theta \right)$, where n is any natural number, all the trigonometric functions remain the same.
For the angles in the form of $\left( \left( 2n-1 \right)\dfrac{\pi }{2}+\theta \right)$, where n is any natural number, sin becomes cos and cos becomes sin, tan becomes cot and cot becomes tan, sec becomes cosec and cosec becomes sec.