prove that sin square x + sin square (x+π/3)+sin square (x-π/3=3/2
Answers
Answer:
Sin²x + Sin²(x + π/3) + Sin²(x - π/3) = 3/2
Step-by-step explanation:
Prove that sin square x + sin square (x+π/3)+sin square (x-π/3=3/2
Sin²x + Sin²(x + π/3) + Sin²(x - π/3)
Using Sin(A + B) + SinACosB + CosASinB
& Sin(A - B) + SinACosB - CosASinB
=Sin²x + (SinxCosπ/3 + CosxSinπ/3)² + (SinxCosπ/3 - CosxSinπ/3)²
using Cosπ/3 = 1/2 Sinπ/3 = √3/2
= Sin²x + (Sinx/2 + Cosx√3/2)² + (Sinx/2 - Cosx√3/2)²
= Sin²x + Sin²x/4 + 3Cos²x/4 + √3SinxCosx/2 + Sin²x/4 + 3Cos²x/4 -√3SinxCosx/2
= Sin²x + Sin²x/4 + 3Cos²x/4 + Sin²x/4 + 3Cos²x/4
= 6Sin²x/4 + 6Cos²x/4
= 3Sin²x/2 + 3Cos²x/2
= (3/2) (Sin²x + Cos²x)
= 3/2
= RHS
QED
Proved
Sin²x + Sin²(x + π/3) + Sin²(x - π/3) = 3/2
Consider the provided information.
\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B
Consider the LHS.
\sin^2A\cos^2B-\cos^2A\sin^2B
\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B (∴\cos^2x=1-\sin^2x)
\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B
\sin^2A-\sin^2B
Hence, proved.