miscellaneous trigonometry question 8 cbse ncert 11th mathematics

miscellaneous trigonometry question 8 cbse ncert 11th mathematics

8.Find  sin(x/2) , cos(x/2) and tan(x/2) if tanx = (-4/3)    x is in the second quadrant.

since x is in the second quadrant

(pi/2) < x < pi so that (pi/4) < (x/2) < (pi/2)

which means that (x/2) is in the first quandrant
and sin(x/2) , cos(x/2) and tan(x/2) are all positive.

given  tanx = (-4/3)
draw a right triangle with x as one of the acute angles
and side opposite to x as 4 units and side adjacent to x is 3 units
using pythagoras theorem hypotenuse = sqrt(16 + 9) = sqrt(25) = 5

since x is in the second quadrant  cosx is negative

cosx = adj / hyp

cosx = (-3/5)

using trigonometry formula trigonometry identities

[sin(x/2)]^2 = (1 - cosx) / 2

[cos(x/2)]^2 = (1 + cosx) / 2

 [sin(x/2)]^2 = [1 - (-3/5) ]  / 2 =[8/5] / 2 = 4/5

[cos(x/2)]^2 = [1 + (-3/5) ]  / 2 =[2/5] / 2 =1/5


take square root and use the fact that (x/2) is in the first quandrant
and sin(x/2) , cos(x/2) and tan(x/2) are all positive.

sin(x/2) = 2 /sqrt(5)

cos(x/2) = 1/sqrt(5)

tan(x/2) = [sin(x/2)]  / [ cos(x/2) ] = [2 /sqrt(5)] /[1 /sqrt(5)] = 2


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7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
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8.Find  sin(x/2) , cos(x/2) and tan(x/2) if tanx = (-4/3)    x is in the second quadrant.
solution