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Anonymous asked in Science & MathematicsMathematics · 9 years ago

# Prove that the LHS equals the RHS: (sinx + sin2x) / (1 + cosx + cos2x) = tanx?

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sin x + sin 2x = sin x + 2 sin x cos x = sin x (1 + 2 cos x)

1 + cos x + cos 2x = 1 + cos x + (2cos^2 x - 1) = cos x + 2 cos^2 x

= cos x (1 + 2 cos x)

∴ (sin x + sin 2x)/(1 + cos x + cos 2x)

= sin x(1 + 2 cos x)/[cos x(1 + 2 cos x)]

= sin x/cos x = tan x.

Note that the double-angle identities for sine and cosine were used:

sin 2x = 2 sin x cos x

cos 2x = 2 cos^2 x - 1.

• (sinx+sin2x) / (1+cosx + cos2x) = tanx

cosx(sinx+sin2x) = sinx(1+cosx+cos2x)

cosx(sinx+2sinxcosx) = sinx(1+cosx +(2cos^2x-1)

sinxcosx(1+2cosx) = sinx(cosx+2cos^2x) = sinxcosx(1+2cosx)

LHS=RHS

• (sin(x) + sin(2x))/(1 + cos(x) + cos(2x)) = tan(x)

Simplifying LHS:

(sin(x))(1 + 2*cos(x))/(1 + cos(x) + cos²(x) - sin²(x)) = (sin(x))(1 + 2*cos(x))/(2*cos²(x) + cos(x))

= sin(x)/cos(x) = tan(x)

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