Question

请数学高手入来

Answer 1

1) 若x+y=90，化简(sin^2y(sinxcosy+sinycosx))/(1-sin^2x) [sin^2y(sinxcosy+sinycosx)]/cos^2x =[sin^2y sin(x+y)]/cos^2x =[sin^2y sin90]/cos^2x =sin^2y/cos^2x ------------------------------------------------------------------------ 2)已知cosx=7/9及x是一个锐角 (a) 求sinx和tanx的值。 ∵sin^2x+cos^2x=1 cosx=7/9 x是一个锐角 ∴sin^2x+(7/9)^2=1 sin^2x=1-(7/9)^2 sin^2x=32/81 sinx=(4√2)/9 ∵sinx=(4√2)/9 cosx=7/9 ∴tanx=sinx/cosx =[(4√2)/9]/( 7/9) =(4√2)/7 (b)由此，求(63(cos^2x-sin^2x))/(16tan^2(90-x)) 以根式表示。 [63(cos^2x-sin^2x)]/[16tan^2(90-x)] =[63(cos2x)]/[16*(1/tanx)^2] =[63(2cos^2x-1)]/(16/tan^2x) ={63[2*(7/9)^2-1]}/{16/[(4√2)/7]^2} =(119/9)/(49/2) =34/63
1) =cos^2 x(sin^2 x+cos^2x)/cos^2x=1 2(a)sinx=4开方(2/9) tanx= 4开方2/7 (b)306/343
参考: my clas *** ate