Question

If $\cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =\lambda \sin \dfrac{\theta }{2}\sin \theta \cos \dfrac{7}{2}\theta$ then $\lambda$ is equals to ?

Answer 1

First you need to simplify the left side of the equation using the cosA - cosB formula two times. Then you get the answer in terms of sinA – sinB . So now, you have to use the formula for sinA – sinB to simplify it and get the answer in terms of sin and cos . Now the left hand side resembles the equation on the right hand side. So, now you just have to compare the values on both sides to get the value of $\lambda$.

Complete step by step solution:
Here is the step wise solution.
The first step we need to solve is to write the left hand side of the equation in terms of the terms on the right hand side of the equation. So, first we use the formula for cosA - cosB two times, to simplify the left hand side. The formula for cosA - cosB is
$\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
So therefore, using this formula we get
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =\left( \cos 2\theta -\cos 3\theta \right)-\left( \cos 4\theta -\cos 5\theta \right)$
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =-2\sin \left( \dfrac{2\theta +3\theta }{2} \right)\sin \left( \dfrac{2\theta -3\theta }{2} \right)+2\sin \left( \dfrac{4\theta +5\theta }{2} \right)\sin \left( \dfrac{4\theta -5\theta }{2} \right)$
Now ,we simplify it to get
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =-2\sin \left( \dfrac{5\theta }{2} \right)\sin \left( \dfrac{-\theta }{2} \right)+2\sin \left( \dfrac{9\theta }{2} \right)\sin \left( \dfrac{-\theta }{2} \right)$
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =2\sin \left( \dfrac{\theta }{2} \right)\left( \sin \left( \dfrac{5\theta }{2} \right)+2\sin \left( \dfrac{9\theta }{2} \right) \right)$
Now we have to use the formula for sinA – sinB to simplify it and get the answer in terms of sin and cos. The formula for sinA – sinB is

$\cos A-\cos B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =4\sin \left( \dfrac{\theta }{2} \right)\left( \cos \left( \dfrac{\dfrac{5\theta }{2}+\dfrac{9\theta }{2}}{2} \right)\sin \left( \dfrac{\dfrac{5\theta }{2}-\dfrac{9\theta }{2}}{2} \right) \right)$
We have to simplify it further to get the answer in terms of the terms in the right hand side.
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =4\sin \left( \dfrac{\theta }{2} \right)\left( \cos \left( \dfrac{7\theta }{2} \right)\sin \left( -\theta \right) \right)$
$\Rightarrow \cos 2\theta -\cos 3\theta -\cos 4\theta +\cos 5\theta =-4\sin \dfrac{\theta }{2}\cos \dfrac{7\theta }{2}\sin \theta$
Now we can compare both the left hand side and the right hand side.
$\Rightarrow -4\sin \dfrac{\theta }{2}\cos \dfrac{7\theta }{2}\sin \theta =\lambda \sin \dfrac{\theta }{2}\sin \theta \cos \dfrac{7}{2}\theta$
We can clearly see that $\lambda =-4$.
Therefore, as we can see , we get the final answer for the question as -4.

So, the correct answer is “Option B”.

Note: You have to remember the trigonometric properties and identities properly to solve these questions. You can remember them easily by solving many trigonometric questions. This way you will get to practice more problems and also remember the formulas.