Question

Simplify and find the value of the following expression $\dfrac{\sin 3A-\cos \left( \dfrac{\pi }{2}-A \right)}{\cos A+\cos \left( \pi +3A \right)}$

Answer 1

Hint: We need to find the value of $\dfrac{\sin 3A-\cos \left( \dfrac{\pi }{2}-A \right)}{\cos A+\cos \left( \pi +3A \right)}$ . We start to solve the given question by finding out the value of $\cos \left( \dfrac{\pi }{2}-A \right)\text{ and cos}\left( \pi +3A \right)$ . Then we will simplify the trigonometric expression using trigonometric formulae to get the desired result.

Complete step by step solution:
We are given a trigonometric expression and are asked to simplify it. We will be solving the given question by finding out the values of $\cos \left( \dfrac{\pi }{2}-A \right)\text{, cos}\left( \pi +3A \right)$ and simplifying the expression using formulae of trigonometry.

The angle $\left( \dfrac{\pi }{2}-A \right)$ lies in the ${{1}^{st}}$ quadrant and every trigonometric function is positive in the ${{1}^{st}}$quadrant.
So, the value of $\cos \left( \dfrac{\pi }{2}-A \right)$ is also positive in the ${{1}^{st}}$ quadrant.
The cosine function changes to the sine function when the angle is 90 or 270.
Following the same, we get,
$\Rightarrow \cos \left( \dfrac{\pi }{2}-A \right)=\sin A$
The angle $\left( \pi +3A \right)$ lies in the ${{3}^{rd}}$ quadrant and only tangent, cotangent is positive in the ${{3}^{rd}}$ quadrant.
So, the value of $\cos \left( \pi +3A \right)$ is negative in the ${{3}^{rd}}$ quadrant.
The cosine function remains the same when the angle is 180 or 360.
Following the same, we get,
$\Rightarrow \cos \left( \pi +3A \right)=-\cos 3A$
Now,
$\Rightarrow \dfrac{\sin 3A-\cos \left( \dfrac{\pi }{2}-A \right)}{\cos A+\cos \left( \pi +3A \right)}$
Substituting the values in the given trigonometric expression, we get,
$\Rightarrow \dfrac{\sin 3A-\sin A}{\cos A-\cos 3A}$
From trigonometry, we know that
$\Rightarrow \sin 3A=3\sin A-4{{\sin }^{3}}A$
Substituting the value of sin3A, we get,
$\Rightarrow \dfrac{3\sin A-4{{\sin }^{3}}A-\sin A}{\cos A-\cos 3A}$
From trigonometry, we know that
$\Rightarrow \cos 3A=4{{\cos }^{3}}A-3\cos A$
Substituting the value of cos3A, we get,
$\Rightarrow \dfrac{3\sin A-4{{\sin }^{3}}A-\sin A}{\cos A-\left( 4{{\cos }^{3}}A-3\cos A \right)}$
Multiplying the minus sign in to the brackets, we get,
$\Rightarrow \dfrac{3\sin A-4{{\sin }^{3}}A-\sin A}{\cos A-4{{\cos }^{3}}A+3\cos A}$
Simplifying the above expression, we get,
$\Rightarrow \dfrac{2\sin A-4{{\sin }^{3}}A}{4\cos A-4{{\cos }^{3}}A}$
Taking 2sinA common from the numerator, we get,
$\Rightarrow \dfrac{2\sin A\left( 1-2{{\sin }^{2}}A \right)}{4\cos A-4{{\cos }^{3}}A}$
Taking 4cosA common from the denominator, we get,
$\Rightarrow \dfrac{2\sin A\left( 1-2{{\sin }^{2}}A \right)}{4\cos A\left( 1-{{\cos }^{2}}A \right)}$
Substituting the same, we get,
$\Rightarrow \dfrac{1}{2}\dfrac{\sin A\left( 1-2{{\sin }^{2}}A \right)}{\cos A\left( 1-{{\cos }^{2}}A \right)}$
From trigonometry,
$\Rightarrow 1-2{{\sin }^{2}}A=\cos 2A$
$\Rightarrow 1-{{\cos }^{2}}A={{\sin }^{2}}A$
Substituting the above values, we get,
$\Rightarrow \dfrac{1}{2}\dfrac{\sin A\times \cos 2A}{\cos A\times {{\sin }^{2}}A}$
Simplifying the above expression, we get,
$\Rightarrow \dfrac{\cos 2A}{2\sin A\cos A}$
From trigonometry, we know that $\sin 2A=2\sin A\cos A$
Substituting the same, we get,
$\Rightarrow \dfrac{\cos 2A}{\sin 2A}$
We know that $\dfrac{\cos A}{\sin A}=\cot A$
Following the same,
$\therefore \dfrac{\sin 3A-\cos \left( \dfrac{\pi }{2}-A \right)}{\cos A+\cos \left( \pi +3A \right)}=\cot 2A$

Note: We must remember that the cosine function is positive in the ${{1}^{st}}$ and ${{4}^{th}}$ quadrants and is negative in the ${{2}^{nd}}$ and ${{3}^{rd}}$ quadrants. The cosine function changes to the sine function when the angle is 90 or 270 and remains the same when the angle is 180 or 360.
$\Rightarrow \cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta$
$\Rightarrow \cos \left( \pi +\theta \right)=\cos \theta$