Show that: (1/2) + cos(x) + cos(2x) + ... + cos(nx) = ((sin(n + (1/2))x)/(2 sin((1/2))x)) By principle of mathematical induction


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 60155 by Tawa1 last updated on 18/May/19

$Show that :$ $\frac{1}{2} + \cos (x) + \cos (2 x) + . . . + \cos (nx) = \frac{\sin (n + \frac{1}{2}) x}{2 \sin (\frac{1}{2}) x}$ $By principle of mathematical induction$
	Answered by Kunal12588 last updated on 18/May/19

	$P (n) : \frac{1}{2} + \cos (x) + \cos (2 x) + . . . + \cos (nx) = \frac{\sin (n + \frac{1}{2}) x}{2 \sin (\frac{1}{2}) x}$ $P (1) :$ $L H S = \frac{1}{2} + c o s (x) = \frac{1 + 2 c o s (x)}{2} = \frac{1 + 2 (1 - 2 {s i n}^{2} \frac{x}{2})}{2}$ $= \frac{3 - 4 {s i n}^{2} \frac{x}{2}}{2}$ $R H S = \frac{s i n (1 + \frac{1}{2}) x}{2 s i n (\frac{1}{2}) x} = \frac{3 s i n (\frac{1}{2}) x - 4 {s i n}^{3} (\frac{1}{2}) x}{2 s i n (\frac{1}{2}) x} = \frac{3 - 4 {s i n}^{2} \frac{x}{2}}{2}$ $\Rightarrow L H S = R H S$ $∴ P (1) i s t r u e$ $L e t P (k) b e t r u e f o r k \in N$ $P (k) \frac{1}{2} + c o s (x) + c o s (2 x) + . . . + c o s (k x) = \frac{s i n (k + \frac{1}{2}) x}{2 s i n (\frac{1}{2}) x} (1)$ $N o w$ $P (k + 1) : \frac{1}{2} + c o s (x) + c o s (2 x) + . . . + c o s (k x) + c o s ((k + 1) x)$ $= \frac{s i n (k + \frac{1}{2}) x}{2 s i n (\frac{1}{2}) x} + c o s ((k + 1) x) [f r o m 1]$ $= \frac{s i n (k + \frac{1}{2}) x + 2 s i n (\frac{1}{2}) x c o s ((k + 1) x)}{2 s i n (\frac{1}{2}) x}$ $= \frac{s i n (k + \frac{1}{2}) x + s i n (k + 1 + \frac{1}{2}) x + s i n (\frac{1}{2} - k - 1) x}{2 s i n (\frac{1}{2}) x}$ $= \frac{s i n (k + \frac{1}{2}) x + s i n (k + 1 + \frac{1}{2}) x - s i n (k + \frac{1}{2}) x}{2 s i n (\frac{1}{2}) x}$ $= \frac{s i n (k + 1 + \frac{1}{2}) x}{2 s i n (\frac{1}{2}) x}$ $∴ P (n) i s t r u e f r o m P r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n$
	Commented by Tawa1 last updated on 18/May/19

	$God bless you sir$
	Commented by Tawa1 last updated on 18/May/19

	$Thanks for your time$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum