Question

what is the value of sin(2A+2B)​

Answer 1

Explanation:

Sum of the angle of a triangle is equal to 180

∘

∠A+∠B+∠C=180

∘

∠B+∠C=180

∘

−∠A

sin(B+C)=sin(180

∘

−A)=sinA

Use the trigonometry identities, sinC+sinD=2sin

2

C+D

cos

2

C−D

So,

Sin2A+sin2B+sin2C=sin2A+2sin(

2

2×(B+C)

)cos(

2

2×(B−C)

)

=sin2A+2sin(B+C)cos(B−C)=sin2A+2sinAcos(B−C)

We know that sin2A=2sinA×cosA

Therefore,

=2sinAcosA+2sinAcos(B−C)=2sinA(cosA+cos(B−C))=2sinA(cos(180

∘

−(B+C))+cos(B−C))

=2sinA(−cos(B+C)+cos(B−C))

We know that cos(A+B)=cosAcosB−sinAsinB and cos(A−B)=cosAcosB+sinAsinB

=2sinA(cosAcosB+sinAsinB−(cosAcosB−sinAsinB))=4sinA×sinB×sinC

Similarly we can do for sin2a+2b