Question

How do you evaluate $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$ ?
(a) Using trigonometric double angle identities
(b) Using linear formulas
(c) a and b both
(d) none of the above

Answer 1

In this problem we are to find the value of $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$. We will try to use the trigonometric double angle identities of $\sin 2\theta$and $\cos 2\theta$ to find and simplify the value of our needed problem. We can start with considering, $\theta =\dfrac{\pi }{7}$ to get ahead with the problem and evaluate the value.

Complete step by step solution:
According to the question, we start with, $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$
Using double angle identities,$\sin \left( 2x \right)=2\sin x\cos x$and $\cos 2x=2{{\cos }^{2}}x-1$
We are to find the value of, $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$
Now, let us consider, $\theta =\dfrac{\pi }{7}$ ,
Then, we have, $2\theta =\dfrac{2\pi }{7}$ ,
So, we can say, $\sin 2\theta =\sin \dfrac{2\pi }{7}$ and $\cos 2\theta =\cos \dfrac{2\pi }{7}$.
And as we can also see, $\dfrac{\pi }{7}$ lies in the first quadrant as $0<\dfrac{\pi }{7}<\dfrac{\pi }{2}$ , where the value of all functions are always positive.
Now,
$\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$
As, $\sin 2\theta =\sin \dfrac{2\pi }{7}$ and $\cos 2\theta =\cos \dfrac{2\pi }{7}$
So, we can also write, $\dfrac{\sin 2\theta }{1+\cos 2\theta }$
Now, again, as said in the trigonometric identities, $\sin \left( 2x \right)=2\sin x\cos x$and $\cos 2x=2{{\cos }^{2}}x-1$
Putting the values,
$\Rightarrow \dfrac{2\sin \theta \cos \theta }{2{{\cos }^{2}}\theta -1+1}$
Cancelling out the 1’s in the denominator,
$\Rightarrow \dfrac{2\sin \theta \cos \theta }{2{{\cos }^{2}}\theta }$
Simplifying, as per the trigonometric identities,
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta$
Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.
Now, as $\theta =\dfrac{\pi }{7}$, we can conclude, $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}=\tan \theta =\tan \dfrac{\pi }{7}$

So, the correct answer is “Option a”.

Note: In the given problem, to find the value of $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1+\cos \left( \dfrac{2\pi }{7} \right)}$ we have used the double angle formulas of trigonometry. Special cases of the sum and difference formulas for sine and cosine yields what is known as the double‐angle identities and the half‐angle identities. Note that there are three forms for the double angle formula for cosine. You only need to know one, but be able to derive the other two from the Pythagorean formula$\left[ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right]$ .