Trigonometry (Continued)


   
 
Theorem 1
General solution of sin q = k
 
 
Method I
 
Given sin q = k .......(i)
 
Let a be any angle, such that sin a = k.
 
(In actual practice, we choose the smallest positive angle such that sin a = k.)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
From (i) and (ii), we have
 
 
Method II
 
Given equation sin q = k.
 
Let a be the least positive angle, such that sin a = k.
 
 
 
Also, any angle co-terminal with the angles a, p - a is trigonometrically equivalent and so has the same 'sine'.
 
 
 
 
 
Particular Cases
 
i) If sin q = 0, then a = 0.
 
Hence, we get q = np + (-1)n 0 = np.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
     
   
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