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Grade:10

4 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

2sinA=2-cosA,1-sin^2(A)=4sin^2(A)-8sinA+4,5sin^2(A)-8sinA+3=0,sinA=(8±2)/10,sinA=1 or 3/5

Ram Avtar
14 Points
9 years ago

Thanks a tonne sir for the solution, I got an idea but could not understand how did you reach the equation SinA=(8+2)/10 from 5sin^2A-8sinA+3.

wont simple factorization by grouping do?  Or it is the latest formulae to obtain the value, if so, kindly illustrate the formulae.

xyz xz
37 Points
9 years ago

  5 sin2A - 8 sinA + 3 { in the form of ax2+bx+c=0 }

 which has the roots

 

x=\frac{-b \pm \sqrt {b^2-4ac}}{2a},

so sinA= -(-8) + (82-4.5.3)1/2           -(-8) - (82-4.5.3)1/2 
                     ---------------------------------------  or ------------------------ 
                               2(5)                               2(5)

           =  {8 + 2}/10 or  {8-2}/10



Saurav Sinha
18 Points
9 years ago

Thanx for the answer sir but can you please elaborate the 1st three steps, i will be really gratefull.

Thanx..

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