Question

If A+B+C =90
then cos2A + cos2B + cos2C =
​

Answer 1

Answer:

90

Step-by-step explanation:

formula of trigonometry identities of class 11

Answer 2

I hope you know the basics of trigonometry. And of course there are many ways to solve a trigonometric problem and this is one of them.

the objective is to simplify

(SinA)^2 + (SinB)^2 + (SinC)^2 (I hope you meant this question only and NOT sin2A + sin2B + sin 2C)

Because it is difficult to directly differentiate a 3 variable function. We must bring it into simplfied form.

and the 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.