本章是高中数学三角函数的最重要内容，是每年高考的必考内容。题目一般是一到两道选择题以及一道解答题的某个分支，分值大多在8到13分之间，难度一般为中低等级。随着新课程标准的实施，对这部分内容的要求有一定的降低倾向，突出“和、差、倍角公式”的作用，突出对正余弦函数的图像与性质的考察。由于新课程标准中向量的引入，将它和平面向量结合起来考察也是高考的一个重要方面。
学习本章，需要熟练背诵本章的所有公式，并且需要熟练地正用、逆用、变形用其中的公式。为了要达到这个目标，需要大量做题，熟练运用公式，熟能生巧，方可学好此章。
两角差的余弦公式
cos ( α − β ) = cos α cos β + sin α sin β {\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }
sin ( α ± β ) = sin α cos β ± cos α sin β {\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }
cos ( α ± β ) = cos α cos β ∓ sin α sin β {\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }
tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β {\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}
sin 2 θ = 2 sin θ cos θ {\displaystyle \sin 2\theta =2\sin \theta \cos \theta }
cos 2 θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ {\displaystyle \cos 2\theta =\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta }
tan 2 θ = 2 tan θ 1 − tan 2 θ {\displaystyle \tan 2\theta ={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}
sin 2 α = 2 tan α 1 + tan 2 α {\displaystyle \sin 2\alpha ={\frac {2\tan \alpha }{1+\tan ^{2}\alpha }}}
cos 2 α = 1 − tan 2 α 1 + tan 2 α {\displaystyle \cos 2\alpha ={\frac {1-\tan ^{2}\alpha }{1+\tan ^{2}\alpha }}}
tan 2 α = 2 tan α 1 − tan 2 α {\displaystyle \tan 2\alpha ={\frac {2\tan \alpha }{1-\tan ^{2}\alpha }}}
sin α cos β = sin ( α + β ) + sin ( α − β ) 2 {\displaystyle \sin \alpha \cos \beta ={\frac {\sin(\alpha +\beta )+\sin(\alpha -\beta )}{2}}}
cos α sin β = sin ( α + β ) − sin ( α − β ) 2 {\displaystyle \cos \alpha \sin \beta ={\frac {\sin(\alpha +\beta )-\sin(\alpha -\beta )}{2}}}
cos α cos β = cos ( α + β ) + cos ( α − β ) 2 {\displaystyle \cos \alpha \cos \beta ={\frac {\cos(\alpha +\beta )+\cos(\alpha -\beta )}{2}}}
sin α sin β = cos ( α − β ) − cos ( α + β ) 2 {\displaystyle \sin \alpha \sin \beta ={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}}}
sin α + sin β = 2 sin α + β 2 cos α − β 2 {\displaystyle \sin \alpha +\sin \beta =2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}
sin α − sin β = 2 cos α + β 2 sin α − β 2 {\displaystyle \sin \alpha -\sin \beta =2\cos {\frac {\alpha +\beta }{2}}\sin {\frac {\alpha -\beta }{2}}}
cos α + cos β = 2 cos α + β 2 cos α − β 2 {\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}
cos α − cos β = − 2 sin α + β 2 sin α − β 2 {\displaystyle \cos \alpha -\cos \beta =-2\sin {\frac {\alpha +\beta }{2}}\sin {\frac {\alpha -\beta }{2}}}
a sin α + b cos α = a 2 + b 2 ⋅ sin ( α + arctan b a ) {\displaystyle a\sin \alpha +b\cos \alpha ={\sqrt {a^{2}+b^{2}}}\cdot \sin \left(\alpha +\arctan {\frac {b}{a}}\right)}
