Question

If A+B+C=90
Then
${sin}^{2} a + {sin}^{2} b + {sin}^{2} c =$
A. 1+4sinAsinBsinC
B. 1-2sinAsinBsinC
C. 2+2sinAsinBsinC
D. 4sinAsinBcosC​

Answer 1

Answer:

condition is A + B + C = 90°

Now apply this formula (SinA)^2 = (1 - Cos2A)/2, then

= (1/2) (1 - Cos2A + 1 - Cos2B + 1 - Cos2C)

= (1/2) (3 - (Cos2A + Cos2B + Cos2C)) ..... (i)

Now apply these formulae

Cos2A + Cos2B = 2 cos((2A+2B)/2) cos((2A-2B)/2) =2 cos(A+B) cos(A-B)

and

Cos2C = 1 - 2 (sinC)^2

Also A+B = 90° - C

so cos(A+B) = SinC

Substituting all these conditions in Cos2A + Cos2B + Cos2C we get

= 2 cos(A+B) cos(A-B) + 1 - 2 (sinC)^2

= 2 sinC cos(A-B) + 1 - 2 (sinC)^2

= 1 + 2 sinC ( cos(A-B) - sin C )

= 1 + 2 sinC ( cos(A-B) - cos(A+B) )

= 1 + 2 sinC (2 sinA sinB) ( We all know this standard formula)

= 1 + 4 sinA sinB sinC .....(ii)

substitute (ii) in (i) then we get

= (1/2) (3 - (1 + 4 sinA sinB sinC))

= (1/2) (2 - 4 sinA sinB sinC)

= 1 - 2 sin A sin B sin C

And that's the required simplified form.

So if A + B + C = 90° then,

(SinA)^2 + (SinB)^2 + (SinC)^2 = 1 - 2 sin A sin B sin C.