Trygonometria 6latek: Dla dwóch kątów jest 1^*) sin(α+β)=sin(α)*cos(β)+cos(α)*sin(β) 2^*) cos(α+β)=cos(α)*cos(β)−sin(α)*sin(β) Mam dojść do takiej postaci 3^*) sin(α+β)=cos(α)*cos(β)*[tg(α)+tg(β)] 4^*) cos(α+β)=cos(α)*cos(β)*[1−tg(α)*tg(β)]

6latek: Teraz dla trzech kątów 5^*) sin(α+β+γ)=sin[(α+β)+γ]= =sin(α)*cos(β)*cos(γ)+cos(α)*sin(β)*cos(γ)+cos(α)*cos(β)*sin(γ)−sin(α)*sin(β)*sin(γ) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 6^*) cos(α+β+γ)=cos[(α+β)+γ]= =cos(α)*cos(β)*cos(γ)−sin(α)*sin(β)*cos(γ)−sin(α)*cos(β)*sin(γ)−cos(α)*sin(β)*sin(γ) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Mam dojśc z tego do postaci 7^*) sin(α+β+γ)=cos(α)*cos(β)*cos(γ)*[tg(α)+tg(β)+tg(γ)−tg(α)*tg(β)*tg(γ)] 8^*) cos(α+β+γ)=cos(α)*cos(β)*cos(γ)*[1−(tg(α)*tg(β)+tg(α)*tg(γ)+tg(β*tg(γ))]