Question

Find the value of
$$\cos (-{89}^o)+\cos (-{87}^o)+\cos (-{85}^o)+....+\cos \left({85}^o\right)+\cos \left({87}^o\right)+\cos \left({89}^o\right)$$

Answer 1

Hint: First of all, use $$\cos (-\theta )=\cos \theta$$ to simplify the equation. Now multiply and divide the equation by $$\sin 1^o$$, take it inside the bracket, and use 2cos A sin B = sin (A + B) – sin (A – B). Now, cancel the like terms and use $$\sin {90}^o=1\text{ and }\sin 0^o=0$$ and finally use $${\displaystyle \frac{1}{\sin \theta }}=\mathrm{cosec}\theta$$ to get the required answer.

Complete step-by-step solution -
In this question, we have to find the value of
$$\cos (-{89}^o)+\cos (-{87}^o)+\cos (-{85}^o)+....+\cos \left({85}^o\right)+\cos \left({87}^o\right)+\cos \left({89}^o\right)$$
Let us consider the expression given in the question in the below format.
$$\begin{array}{rl}& E=\cos (-{89}^o)+\cos (-{87}^o)+\cos (-{85}^o)+....+\cos (-3^o)+\cos (-1^o)+\cos \left(1^o\right)+\\ & \cos \left(3^o\right)+......+\cos \left({85}^o\right)+\cos \left({87}^o\right)+\cos \left({89}^o\right)\end{array}$$
By adding one term from the end and one term from the beginning, the second term from the end and second term from the beginning and keep repeating this, we get,
$$E=2(\cos {89}^o+\cos {87}^o+\cos {85}^o+\cos {83}^o+....+\cos 5^o+\cos 3^o+\cos 1^o)$$
Now, multiplying $$\sin 1^o$$ on both numerator and denominator of the above expression, we get,
$$E=2{\displaystyle \frac{\sin 1^o}{\sin 1^o}}(\cos 1^o+\cos 3^o+\cos 5^o+......+\cos {87}^o+\cos {89}^o)$$
We can also write the above equation as,
$$E={\displaystyle \frac{1}{\sin 1^o}}(2\cos 1^o\sin 1^o+2\cos 3^o\sin 1^o+2\cos 5^o\sin 1^o+......+2\cos {87}^o\sin 1^o+2\cos {89}^o\sin 1^o)$$
We know that, 2 cos A sin B = sin (A + B) – sin (A – B). By using this in the above expression, we get,
$$E={\displaystyle \frac{1}{\sin 1^o}}\left[\begin{array}{rl}& \sin (1^o+1^o)-\sin (1^o-1^o)+\sin (3^o+1^o)-\sin (3^o-1^o)+\sin (5^o+1^o)-\sin (5^o-1^o)......\\ & +\sin ({87}^o+1^o)-\sin ({87}^o-1^o)+\sin ({89}^o+1^o)-\sin ({89}^o-1^o)\end{array}\right]$$
$$E={\displaystyle \frac{1}{\sin 1^o}}[\sin 2^o-\sin 0^o+\sin 4^o-\sin 2^o+\sin 6^o-\sin 4^o......+\sin {88}^o-\sin {86}^o+\sin {90}^o-\sin {88}^o]$$
By canceling the like terms, we finally get,
$$E={\displaystyle \frac{1}{\sin 1^o}}[-\sin 0^o+\sin {90}^o]$$
We know that $$\sin {90}^o=1\text{ and }\sin 0^o=0$$. By substituting these values, we get,
$$E={\displaystyle \frac{1}{\sin 1^o}}[-0+1]$$
$$E={\displaystyle \frac{1}{\sin 1^o}}$$
We know that $${\displaystyle \frac{1}{\sin \theta }}=\mathrm{cosec}\theta$$.
So, we get,
$$E=\mathrm{cosec}{1}^o$$
So, we get the value of the given expression as $$\mathrm{cosec}{1}^o$$.

Note: In this question, first of all, students often make the mistake of writing $$\cos (-\theta )\text{ as }-\cos \theta$$ which is wrong because $$\cos (-\theta )=\cos \theta$$. Now apart from multiplying $$\sin 1^o$$, students often try to club $${89}^o\text{ with }{1}^o,{87}^o\text{ with }{3}^o$$ and so on. They are complementary angles but that leads to lengthy solutions or they often get stuck. And always try to convert the product into different formulas to cancel the terms.