Question

If α and β are the angles in the first quadrant tan α=1/7 sin β=1/√10 then using the formula sin(α β)=sinα cosβ cosα sinβ find the value of α+2β

Answer 1

 given  0 <= A or B  <= 90°.  So the trigonometric ratios are all positive,

       tan A = 1/7
          => Sec² A = 1 + 1/49 = 50/49      =>         Cos A = 7/√50
           Sin² A = 1 - 49/50 = 1/50         =>    Sin A = 1/√50

           sin B = 1/√10      => Cos² B  = 1 - 1/10 = 9/10      => Cos B = 3/√10

Sin (A + B) = Sin A cos B + Cos A Sin B 
                 = 3/√500  + 7/√500 =  10/√500  =   1/√5
Cos² (A+B) = 1 - 1/5 = 4/5          =>  Cos (A+B) = 2/√5

Sin (A+B  +  B) = Sin (A+B) Cos B + Cos(A+B) SIn B
                     = 3/√50 + 2/√50 =  1/√2

A + 2 B = 45°
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You may also find the value of  Sin 2B,  Cos 2B  using their formulas.
then  find  Sin  (A+2B ).

Sin 2B = 6/10 = 3/5            Cos2B = 9/10 - 1/10 = 4/5

So Sin(A+2B) = 1/√50 * 4/5  + 3/5 * 7/√50
           = 1/√2
 So A + 2B = 45 deg.