हल कीजिए - (i) ` sin 3 theta = sin 2 theta` (ii) ` sin m theta + sin n theta = 0` (iii) ` tan 3 theta tan theta = 1`

हल कीजिए - (i) ` sin 3 theta = sin 2 theta` (ii) ` sin m

1.	हल कीजिए - (i) ` sin 3 theta = sin 2 theta` (ii) ` sin m theta + sin n theta = 0` (iii) ` tan 3 theta tan theta = 1`
Answer» दिया है - `sin 3 theta = sin 2 theta` `= sin [n pi + (-1)^(n) * 2 theta]` `rArr " " 3 theta = n pi + (-1)^(n) * 2 theta ` यदि n सम है अर्थात n = 2m तब `3 theta = 2 m pi + 2 theta` `rArr theta = 2 m pi` और यदि n विषम संख्या है अर्थात `n = 2m +1`, तब `rArr 3 theta = (2m +1) pi - 2 theta` `rArr 5 theta = (2m +1) pi` `rArr theta = (2m +1) * pi/5` (ii) दिया है - `sin m theta + sin n theta = 0` `2 sin [((m+n)theta)/2] cos [((m-n)theta)/2]= 0` यदि `sin((m+n)/2) theta = 0 rArr ((m+n)/2) theta = k pi, " " k in Z` `rArr theta = (2kpi)/(m+n), " " k in Z` यदि `cos((m-n)/2) theta = 0 `तब `((m-n)/2) theta = (2k+1) pi/2, " " k in Z` `rArr theta = ((2k+1)pi)/(m-n) , " " k in Z` (iii) दिया है - `tan 3 theta tan theta =1` `rArr tan 3 theta = 1/(tan theta) = cot theta = tan (pi/2 - theta)` `rArr 3 theta = n pi + (pi/2 - theta) ` `rArr 4 theta = (2 pi +1) pi/2` ` rArr theta = (2n +1) pi/8`