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tAn x=1/2,sin²(π/4+x)=

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

=(1/2)cos(2x)

令tan(x-π/4)=0 x-π/4=kπ x=π/4+kπ 在(0,2π) x可取,π/4,5π/4, 则,间断点,两个

tan(π/4-x) =1+tan45tanx分之tan45°-tanx答案就是这个。是公式。记得采纳!

sin(π/4+x)sin(π/4-x) =sin[π/2-(π/4-x)]sin(π/4-x) =cos(π/4-x)sin(π/4-x) =1/2sin(π/2-2x) =1/2cos2x=1/6 cos2x=1/3 x在(pai/2,pai),则2x在(pai,2pai) 所以:sin2x=-√[1-(cos2x)^2]=-2√2/3 sin4x=2sin2xcos2x=-2*1/3*2√2/3=-4√2/9 然后再求出co...

因为a为锐角,且tana=(√2)-1 所以tan2a=(2tana)/(1-tan²a)=1 所以sin2a=√2/2,cos2a=√2/2,sin(2a+π/4)=1 所以f(x)=x²+x 因为a(n+1)=f(an) 所以a(n+1)=an²+an=an(1+an) 取倒数, 1/an(1+an)=1/a(n+1) 所以,1/(1+an)=1/a(n)-1/a(n...

cosx/1+sinx =【cos²(x/2)-sin²(x/2)】/【sin²(x/2)+2sin(x/2)cos(x/2)+cos²(x/2)】 =[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]/[cos(x/2)+sin(x/2)]² =[cos(x/2)-sin(x/2)]/[cos(x/2)+sin(x/2)] =√2sin(π/4-x/2)/√2cos...

-√6/3 根据公式:(cosx-sinx)²=1-2sinxcosx=1-1/3=2/3 所以 cosx-sinx=+-√6/3 因为 π/4

因为(π/4-x)+(π/4+x)=π/2,所以tan(π/4-x)=cot(π/4+x) (2cos²x-1)/2tan(π/4-x)sin²(π/4+x) =sin2x/[2cot(π/4+x)sin²(π/4+x)] =sin2x/{[2cos(π/4+x)/sin(π/4+x)]sin²(π/4+x)} =sin2x/[2sin(π/4+x)cos(π/4+x)] =sin2x/sin(π/2...

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