NBlog - 自己推导的三角函数二级结论

做题推导出的一些小结论, 之后可能会继续更新.

1. $\frac{1}{c o s ^{2} α} - 1 = tan^{2} α$

tan α = \frac{sin α}{cos α}

tan^{2} α + 1 = \frac{sin ^{2} α}{cos ^{2} α} + 1

tan^{2} α + 1 = \frac{sin ^{2} α + cos ^{2} α}{cos ^{2} α}

\frac{1}{cos ^{2} α} - 1 = tan^{2} α

2. tan(4π+α)=±1−sin2α1+sin2α<path d="M983 90

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tan (\frac{π}{4} + α)

= \frac{sin ( \frac{π}{4} + α )}{cos ( \frac{π}{4} + α )}

= \frac{cos α + sin α}{cos α - sin α}

= \pm \frac{( cos α + sin α ) ^{2}}{( cos α - sin α ) ^{2}}

= \pm \frac{1 + 2 cos α sin α}{1 - 2 cos α sin α}

= \pm \frac{1 + sin 2 α}{1 - sin 2 α}

3. asinθ+bcosθ=a2+b2<path d="M95,702

c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>sin(θ+ϕ), 其中 $tan ϕ = \frac{b}{a}$

设角 $ϕ$ , 令

tan ϕ = \frac{b}{a}

则有

a sin θ + b cos θ

= a^{2} + b^{2} (\frac{a}{a ^{2} + b ^{2}} sin θ + \frac{b}{a ^{2} + b ^{2}} cos θ)

= a^{2} + b^{2} (cos ϕ sin θ + sin ϕ cos θ)

= a^{2} + b^{2} sin (θ + ϕ)

自己推导的三角函数二级结论

1. cos2α1​−1=tan2α

2. tan(4π​+α)=±1−sin2α1+sin2α​<path d="M983 90

3. asinθ+bcosθ=a2+b2<path d="M95,702

1. $\frac{1}{c o s ^{2} α} - 1 = tan^{2} α$

2. tan(4π+α)=±1−sin2α1+sin2α<path d="M983 90