sin^3a cos^3a=1,sina cosa
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![sin^3a cos^3a=1,sina cosa](/uploads/image/f/793910-38-0.jpg?t=sin%5E3a+cos%5E3a%3D1%2Csina+cosa)
f(x)=-2acos²x-2√2asinx+3a+b=-2a(1-sin²x)-2√2asinx+3a+b=2asin²x-2√2asinx+a+b=2a(sin
1.(sina)^2+(sinb)^2-(sinasinb)^2+(cosacosb)^2=(sina)^2-(sinasinb)^2+1-(cosb)^2+(cosacosb)^2=(sina)^2
(1)tana=-4∴cota=-1/4csca=±√(cot²a+1)=±√17/4sina=±(4/17)√17(2)3sinacosa=(3/2)sin2a万能公式:sin2a=2ta
三角函数证明方法(1)证明一个等式有几种思路:1、从一边到另一边;2、先证明另一个等式成立,从而推出需要证明的等式成立;3、证明左右等于同一个式子;另外三角恒等式证明中要善于用“1”.(2)方法一:消
因sina²+cosα²=1全都平方b²cosα²=a²cosβ²sinα²=a²sinβ²两市相加b&sup
第一问的方法是将1拆成sin²a+cos²a,然后就能算了第二问用到常用的倍角公式:cos2θ=cos²a-sin²a=2cos²a-1=1-2sin
sin²a+cosˇ4a+sin²acos²a=sin²a+cos²a(cos²a+sin²a)=sin²a+cos
cos(2a)=1/4[sin(2a)]^2=1-[cos(2a)]^2=1-1/16=15/16(cosa)^4+(sina)^4+(sina)^2(cosa)^2=[(cosa)^2+(sina)
2SINA-SINACOSA-3COSA=0两边同时除以cosA*cosA2tan^2(A)-tanA-3=0tanA=-1或tanA=3/2.A=-π/4或3π/4或{sinA=3/√13和cosA
原式=cos²a(2-2sin²a+3)-2sin²acos²a-3sin²a-4sin²acos²a+3=5cos²a
正弦定理知等价于证sinacosa+sinbcosb+sinccosc=2sinasinbsin(a+b)=2sin^2asinbcosb+2sin^2bsinacosa移项用二倍角公式等价于cos2
acos^2C/2+ccos^2A/2=3b/2a*(cosC+1)/2+c*(cosA+1)/2=3b/2acosC+a+ccosA+c=3bacosC+a+ccosA+c=2b+b,a/sinA=
y=sin(x+π/6)sin(x-π/6)+acosx=-1/2[cos(x+π/6+x-π/6)-cos(x+π/6-x+π/6)+acosx=-1/2(cos2x-cosπ/3)+acosx=-
∵左边=sin^4+cos^4=(sin^2+cos^2)^2-2sin^2cos^2而sin^2+cos^2=1,∴sin^4+cos^4=1-2sin^2cos^2=右边
∵(1-sin^4a-cos^4a)=1-(sin²a+cos²a)(sin²a-cos²a)=1+cos2a=1+2cos²a-1=2cos&sup
1)f(x)=a[1/2*sin2x-√3/2*(1+cos2x)+√3/2]+b=a[1/2sin2x-√3/2cos2x]+b=asin(2x-π/3)+b因为a>0,所以单调减区间为:2kπ+π
sin^4a+cos+sinacosa=(sin^4a+sinacosa)+cosa=sina(sina+cosa)+cosa=sina+cosa=1,得证!
sin^6α+cos^6α+3sin^2αcos^2α=(sin^2a)^3+(cos^2a)^3+3sin^2αcos^2α=(sin^2a+cos^2a)(sin^4a-sin^2acos^2a+