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Have both axles inspected for excessive wear and also the hub bearings by the wheels. Secondly, have the tires rebalanced. Don't care how old they are. Third, replace the left front strut assembly. The...

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...

Expand the RHS -2[sin(x/2)cos(y/2)+sin(y/2)cos(x/2)][sin(x/2)cos(y/2)-sin(y/2)cos(x/2)]=-2[sin^2(x/2cos^2(y/2)-sin^2(y/2)cos^2(x/2)] using difference of two squares change the cos^2(y/2)=1-sin^2(y/2) and cos^2(x/2)=(1-sin^2x/2) to obtain -2(sin^2(x/2)-sin^2(y/2)) convert the sin^2(x/2) and sin^2(y/2) to the double...

9) LHS = cos ( x + y - y ) LHS = cos... LHS = sin x - ( sin y / cos y )*cos x LHS = (sin x cos y - sin y cos x ) / cos...2 x cos^2 y - sin^2 x sin^2 y ) LHS = ( sin^2 x - sin^2 x sin^2 y - sin...

Doubtful, if this is possible. I would present it on the mathematics category for a firm answer.

LHS = ( sin x + sin 3x) / ( cos x + cos 3x ) LHS = 2 sin ( 2x) cos (x) / 2 cos (2x ) cos ( x ) LHS = tan 2x LHS = RHS QED

Since cos^2(x)=(1+cos2x)/2 so we obtain cos2x=2cos^(x)-1 also we know sin2x=2sin(x)cos(x) now if we put these things in what u have writen we will have [(2sin(x)cos(x)]/sin(x) - (2cos^(x)-1)/cos(x) = 2cos(x)-2cos(x)+ (1/cos(x))=sec(x)