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Question: The value of \[\cos \left( A \right)+\cos \left( A+B \right)+\cos \left( A+2B \right)+.....+\cos \le...

The value of cos⁡(A)+cos⁡(A+B)+cos⁡(A+2B)+.....+cos⁡(A+(n−1)B)=sin⁡(nB2)cos⁡(A+(n−1)B2)sin⁡(B2)\cos \left( A \right)+\cos \left( A+B \right)+\cos \left( A+2B \right)+.....+\cos \left( A+\left( n-1 \right)B \right)=\dfrac{\sin \left( \dfrac{nB}{2} \right)\cos \left( A+\left( n-1 \right)\dfrac{B}{2} \right)}{\sin \left( \dfrac{B}{2} \right)} is
a) True
b) False

Explanation

Solution

Hint: We know that any complex term of the form eiθ{{e}^{i\theta }} can be expressed as the eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta , so, we can write cos⁡θ=Re⁡(eiθ)\cos \theta =\operatorname{Re}\left( {{e}^{i\theta }} \right). After applying this formula, we can apply the formula which is used to find the sum of the geometric series, that is, S=a(rn−1)(r−1)S=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}, where a is the first term of geometric progression, r is the common ratio of the geometric progression.

Complete step-by-step answer:
In this question, we have to find whether cos⁡(A)+cos⁡(A+B)+cos⁡(A+2B)+.....+cos⁡(A+(n−1)B)=sin⁡(nB2)cos⁡(A+(n−1)B2)sin⁡(B2)\cos \left( A \right)+\cos \left( A+B \right)+\cos \left( A+2B \right)+.....+\cos \left( A+\left( n-1 \right)B \right)=\dfrac{\sin \left( \dfrac{nB}{2} \right)\cos \left( A+\left( n-1 \right)\dfrac{B}{2} \right)}{\sin \left( \dfrac{B}{2} \right)} is true or not.
Now, let us consider L.H.S. first, that is, cos⁡(A)+cos⁡(A+B)+cos⁡(A+2B)+.....+cos⁡(A+(n−1)B)\cos \left( A \right)+\cos \left( A+B \right)+\cos \left( A+2B \right)+.....+\cos \left( A+\left( n-1 \right)B \right)
We know that this can be further written as ∑k=0n−1cos⁡(A+kB)\sum\limits_{k=0}^{n-1}{\cos \left( A+kB \right)}
We know that any exponential complex number, like eiθ{{e}^{i\theta }} can be written as eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta , where cos⁡θ\cos \theta is the real part of the complex number and sin⁡θ\sin \theta is the imaginary part. So, cos⁡θ\cos \theta can be written as cos⁡θ=Re⁡(eiθ)\cos \theta =\operatorname{Re}\left( {{e}^{i\theta }} \right).
Therefore, we can write, ∑k=0n−1cos⁡(A+kB)=∑k=0n−1Re⁡(ei(A+kB))\sum\limits_{k=0}^{n-1}{\cos \left( A+kB \right)}=\sum\limits_{k=0}^{n-1}{\operatorname{Re}\left( {{e}^{i\left( A+kB \right)}} \right)}
Now, if we will put the value of k in ∑k=0n−1Re⁡(ei(A+kB))\sum\limits_{k=0}^{n-1}{\operatorname{Re}\left( {{e}^{i\left( A+kB \right)}} \right)}, we will see that a geometric progression has formed, which is as same as Re⁡(ei(A))+Re⁡(ei(A+B))+Re⁡(ei(A+2B))+.....+Re⁡(ei(A+(n−1)B))\operatorname{Re}\left( {{e}^{i\left( A \right)}} \right)+\operatorname{Re}\left( {{e}^{i\left( A+B \right)}} \right)+\operatorname{Re}\left( {{e}^{i\left( A+2B \right)}} \right)+.....+\operatorname{Re}\left( {{e}^{i\left( A+\left( n-1 \right)B \right)}} \right), which on further simplification gives,
Re⁡((ei(A))+(ei(A+B))+(ei(A+2B))+.....+(ei(A+(n−1)B)))\operatorname{Re}\left( \left( {{e}^{i\left( A \right)}} \right)+\left( {{e}^{i\left( A+B \right)}} \right)+\left( {{e}^{i\left( A+2B \right)}} \right)+.....+\left( {{e}^{i\left( A+\left( n-1 \right)B \right)}} \right) \right)
⇒Re⁡(eiA)(1+(ei(B))+(ei(2B))+.....+(ei((n−1)B)))\Rightarrow \operatorname{Re}\left( {{e}^{iA}} \right)\left( 1+\left( {{e}^{i\left( B \right)}} \right)+\left( {{e}^{i\left( 2B \right)}} \right)+.....+\left( {{e}^{i\left( \left( n-1 \right)B \right)}} \right) \right)
Now, we can see that a pattern is followed, which is as same as the pattern of geometric progression, so we can apply the formula of sum of geometric series, that is, S=a(rn−1)(r−1)S=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}, where a represents the first term of geometric progression, r represents the common ratio of the geometric progression.
In our series, 1 is the first term of geometric progression, ei(B){{e}^{i\left( B \right)}} is the common ratio of the progression, that is, a=1a=1 and r=ei(B)r={{e}^{i\left( B \right)}}.
Therefore, the sum of the series (1+(ei(B))+(ei(2B))+.....+(ei((n−1)B)))\left( 1+\left( {{e}^{i\left( B \right)}} \right)+\left( {{e}^{i\left( 2B \right)}} \right)+.....+\left( {{e}^{i\left( \left( n-1 \right)B \right)}} \right) \right) is S=1((ei(B))n−1)((ei(B))−1)S=\dfrac{1\left( {{\left( {{e}^{i\left( B \right)}} \right)}^{n}}-1 \right)}{\left( \left( {{e}^{i\left( B \right)}} \right)-1 \right)}
⇒S=((einB)−1)((ei(B))−1)\Rightarrow S=\dfrac{\left( \left( {{e}^{inB}} \right)-1 \right)}{\left( \left( {{e}^{i\left( B \right)}} \right)-1 \right)}
Therefore, Re⁡(eiA)(1+(ei(B))+(ei(2B))+.....+(ei((n−1)B)))=Re⁡(eiA)(1((einB)−1)((ei(B))−1))\operatorname{Re}\left( {{e}^{iA}} \right)\left( 1+\left( {{e}^{i\left( B \right)}} \right)+\left( {{e}^{i\left( 2B \right)}} \right)+.....+\left( {{e}^{i\left( \left( n-1 \right)B \right)}} \right) \right)=\operatorname{Re}\left( {{e}^{iA}} \right)\left( \dfrac{1\left( \left( {{e}^{inB}} \right)-1 \right)}{\left( \left( {{e}^{i\left( B \right)}} \right)-1 \right)} \right)
Now, we will take, einB2{{e}^{in\dfrac{B}{2}}} common from numerator and eiB2{{e}^{i\dfrac{B}{2}}} common from denominator, we will get,
Re⁡(eiA)(ei(n−1)B2)((einB2)−(e−inB2)(eiB2)−(e−iB2))\operatorname{Re}\left( {{e}^{iA}} \right)\left( {{e}^{i(n-1)\dfrac{B}{2}}} \right)\left( \dfrac{\left( {{e}^{in\dfrac{B}{2}}} \right)-\left( {{e}^{-in\dfrac{B}{2}}} \right)}{\left( {{e}^{i\dfrac{B}{2}}} \right)-\left( {{e}^{-i\dfrac{B}{2}}} \right)} \right)
⇒Re⁡(ei(A+(n−1)B2))((einB2)−(e−inB2)(eiB2)−(e−iB2))\Rightarrow \operatorname{Re}\left( {{e}^{i\left( A+(n-1)\dfrac{B}{2} \right)}} \right)\left( \dfrac{\left( {{e}^{in\dfrac{B}{2}}} \right)-\left( {{e}^{-in\dfrac{B}{2}}} \right)}{\left( {{e}^{i\dfrac{B}{2}}} \right)-\left( {{e}^{-i\dfrac{B}{2}}} \right)} \right)
Now, we will put eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta , so, we will get,
⇒Re⁡(cos⁡(A+(n−1)B2)+isin⁡(A+(n−1)B2))((cos⁡(nB2)+isin⁡(nB2))−(cos⁡(nB2)−isin⁡(nB2))(cos⁡B2+isin⁡B2)−(cos⁡B2−isin⁡B2))\Rightarrow \operatorname{Re}\left( \cos \left( A+(n-1)\dfrac{B}{2} \right)+i\sin \left( A+(n-1)\dfrac{B}{2} \right) \right)\left( \dfrac{\left( \cos \left( n\dfrac{B}{2} \right)+i\sin \left( n\dfrac{B}{2} \right) \right)-\left( \cos \left( n\dfrac{B}{2} \right)-i\sin \left( n\dfrac{B}{2} \right) \right)}{\left( \cos \dfrac{B}{2}+i\sin \dfrac{B}{2} \right)-\left( \cos \dfrac{B}{2}-i\sin \dfrac{B}{2} \right)} \right)
⇒Re⁡(cos⁡(A+(n−1)B2)+isin⁡(A+(n−1)B2))(2(isin⁡(nB2))2(isin⁡B2))\Rightarrow \operatorname{Re}\left( \cos \left( A+(n-1)\dfrac{B}{2} \right)+i\sin \left( A+(n-1)\dfrac{B}{2} \right) \right)\left( \dfrac{2\left( i\sin \left( n\dfrac{B}{2} \right) \right)}{2\left( i\sin \dfrac{B}{2} \right)} \right)
⇒Re⁡(cos⁡(A+(n−1)B2)+isin⁡(A+(n−1)B2))(sin⁡(nB2)sin⁡B2)\Rightarrow \operatorname{Re}\left( \cos \left( A+(n-1)\dfrac{B}{2} \right)+i\sin \left( A+(n-1)\dfrac{B}{2} \right) \right)\left( \dfrac{\sin \left( n\dfrac{B}{2} \right)}{\sin \dfrac{B}{2}} \right)
Now, we will remove the terms containing ii, to get the real part as L.H.S.
⇒(cos⁡(A+(n−1)B2))(sin⁡(nB2)sin⁡B2)\Rightarrow \left( \cos \left( A+(n-1)\dfrac{B}{2} \right) \right)\left( \dfrac{\sin \left( n\dfrac{B}{2} \right)}{\sin \dfrac{B}{2}} \right)
⇒(cos⁡(A+(n−1)B2)×sin⁡(nB2)sin⁡B2)\Rightarrow \left( \dfrac{\cos \left( A+(n-1)\dfrac{B}{2} \right)\times \sin \left( n\dfrac{B}{2} \right)}{\sin \dfrac{B}{2}} \right)
= R.H.S.
Hence, proved.

Note: There are possibilities that we will do rationalisation at the place of taking out common from numerator and denominator, which is obviously not wrong but will make questions more complicated and lengthier.