Преобразование произведения тригонометрических функций в сумму

п.1. Произведение двух тригонометрических функций

Чтобы получить формулы для произведений двух тригонометрических функций, используем промежуточные результаты, полученные в §17 данного справочника.
Введя новые переменные: \(x=\frac{\alpha+\beta}{2},\ y=\frac{\alpha-\beta}{2}\), для \(\alpha=x+y,\ \beta=x-y\) мы получили:

\begin{gather*} sin\alpha+sin\beta=sin(x+y)+sin(x-y)=2sinxcosy\\ cos\alpha+cos\beta=cos(x+y)+cos(x-y)=2cosxcosy\\ cos\alpha-cos\beta=cos(x+y)-cos(x-y)=-2sinxsiny \end{gather*}

п.2. Примеры

Пример 1. Вычислите:
a) \begin{gather*} \frac{1}{cos20^{\circ}}-4sin50^{\circ}=\frac{1-4sin50^{\circ}cos20^{\circ}}{cos20^{\circ}}=\frac{1-4\cdot\frac{sin(50^{\circ}+20^{\circ})+sin(50^{\circ}-20^{\circ})}{2}}{cos20^{\circ}}=\\ =\frac{1-2(sin70^{\circ}+sin30^{\circ})}{cos20^{\circ}}=\frac{1-2sin70^{\circ}-2\cdot\frac12}{cos20^{\circ}}=\frac{-2sin70^{\circ}}{cos(90^{\circ}-20^{\circ})}=\frac{-2sin70^{\circ}}{sin70^{\circ}}=-2 \end{gather*}

б) \begin{gather*} \frac{cos64^{\circ}cos4^{\circ}-cos86^{\circ}cos26^{\circ}}{cos71^{\circ}cos41^{\circ}-cos49^{\circ}cos19^{\circ}}=\frac{\frac{cos68^{\circ}+cos60^{\circ}}{2}-\frac{cos112^{\circ}+cos60^{\circ}}{2}}{\frac{cos112^{\circ}+cos30^{\circ}}{2}-\frac{cos68^{\circ}+cos30^{\circ}}{2}}=\\ =\frac{cos68^{\circ}+cos60^{\circ}-cos112^{\circ}-cos60^{\circ}}{cos112^{\circ}+cos30^{\circ}-cos68^{\circ}+cos30^{\circ}}=\frac{cos68^{\circ}-cos112^{\circ}}{cos112^{\circ}-cos68^{\circ}}=-1 \end{gather*}

в) \begin{gather*} sin20^{\circ}sin70^{\circ}-sin14^{\circ}cos26^{\circ}-cos6^{\circ}cos84^{\circ}=\\ =-\frac{cos(6^{\circ}+70^{\circ})+cos(6^{\circ}-84^{\circ})}{2}=-\frac{cos90^{\circ}-cos50^{\circ}}{2}-\frac{sin40^{\circ}-sin12^{\circ}}{2}-\\ -\frac{cos90^{\circ}+cos78^{\circ}}{2}=\frac{50^{\circ}-sin40^{\circ}+sin12^{\circ}-cos78^{\circ}}{2}=\\ =\frac{cos50^{\circ}-sin(90^{\circ}-50^{\circ})+sin(90^{\circ}-78^{\circ})-cos78^{\circ}}{2}=\\ \frac{cos50^{\circ}-cos50^{\circ}+cos78^{\circ}-cos78^{\circ}}{2}=0 \end{gather*}

г*) \begin{gather*} \frac{1}{cos310^{\circ}}-\frac{1}{\sqrt{3}cos220^{\circ}}=\frac{1}{cos(310^{\circ}-360^{\circ})}-\frac{1}{\sqrt{3}cos(220^{\circ}-360^{\circ})}=\\ =\frac{1}{cos50^{\circ}}-\frac{1}{\sqrt{3}cos140^{\circ}}=\frac{\sqrt{3}cos140^{\circ}-cos50^{\circ}}{\sqrt{3}cos140^{\circ}cos50^{\circ}}=\\ =\frac{\sqrt{3}cos140^{\circ}-cos50^{\circ}}{\sqrt{3}\cdot\frac{cos(140^{\circ}+50^{\circ})+cos(140^{\circ}-50^{\circ})}{2}}=\frac{2(\sqrt{3}cos140^{\circ}-cos50^{\circ})}{\sqrt{3}(cos190^{\circ}+0)}=\\ =\frac{2(\sqrt{3}cos(180^{\circ}-40^{\circ})-cos(90^{\circ}-40^{\circ}))}{\sqrt{3}cos(180^{\circ}+10^{\circ})}=\frac{2(-\sqrt{3}cos40^{\circ}-sin40^{\circ})}{-\sqrt{3}cos10^{\circ}}=\\ =\frac{2(\sqrt{3}cos40^{\circ}+sin40^{\circ})}{\sqrt{3}cos10^{\circ}}=\frac{4\left(\frac{\sqrt{3}}{2}cos40^{\circ}+\frac12sin40^{\circ}\right)}{\sqrt{3}cos10^{\circ}}=\\ =\frac{4(cos30^{\circ}cos40^{\circ}+sin30^{\circ}sin40^{\circ})}{\sqrt{3}cos10^{\circ}}=\frac{4cos(40^{\circ}-30^{\circ})}{\sqrt{3}cos10^{\circ}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3} \end{gather*}

Пример 2.Упростите выражение:
a) \begin{gather*} 2-8cos^2\alpha-8sin\left(\alpha-\frac\pi3\right)sin\left(\alpha+\frac\pi3\right)=\\ =2-8\cdot\frac{1+cos2\alpha}{2}+8\cdot\frac{cos\left(\alpha-\frac\pi3+\alpha+\frac\pi3\right)-cos\left(\alpha-\frac\pi3-\alpha-\frac\pi3\right)}{2}=\\ =2-4(1+cos2\alpha)+4\left(cos2\alpha-cos\frac{2\pi}{3}\right)=\\ =2-4-4cos2\alpha+4cos2\alpha-4\cdot\left(-\frac12\right)=0 \end{gather*}
б) \begin{gather*} cos^2\alpha+cos^2\beta-cos(\alpha+\beta)cos(\alpha-\beta)=\\ =cos^2\alpha+cos^2\alpha-\frac{cos(\alpha+\beta+\alpha-\beta)+cos(\alpha+\beta-\alpha+\beta)}{2}=\\ =\frac{1+cos2\alpha}{2}+\frac{1+cos2\beta}{2}-\frac{cos2\alpha+cos2\beta}{2}=\\ =\frac{2+cos2\alpha+cos2\beta-cos2\alpha-cos2\beta}{2}=1 \end{gather*}
в) \begin{gather*} cos^2\left(\frac{3\pi}{8}+\alpha\right)-sin^2\left(\frac{3\pi}{8}-\alpha\right)=\frac{1+cos\left(2\left(\frac{3\pi}{8}+\alpha\right)\right)}{2}-\frac{1-cos\left(2\left(\frac{3\pi}{8}-\alpha\right)\right)}{2}=\\ =\frac{cos\left(\frac{3\pi}{4}+2\alpha\right)+cos\left(\frac{3\pi}{4}-2\alpha\right)}{2}=cos\frac{3\pi}{4}cos2\alpha=-\frac{\sqrt{2}}{2}cos2\alpha \end{gather*}
г) \begin{gather*} 2sin\left(\frac{\pi}{10}-3\alpha\right)sin\left(3\alpha-\frac{3\pi}{5}\right)-cos\left(6\alpha-\frac{7\pi}{10}\right)=\\ =-\left(cos\left(\frac{\pi}{10}-3\alpha+3\alpha-\frac{3\pi}{5}\right)-cos\left(\frac{\pi}{10}-3\alpha-3\alpha+\frac{3\pi}{5}\right)\right)-cos\left(6\alpha-\frac{7\pi}{10}\right)=\\ =-cos\frac\pi2+cos\left(\frac{7\pi}{10}-6\alpha\right)-cos\left(6\alpha-\frac{7\pi}{10}\right)=0 \end{gather*}