Respuesta :
[tex]\bf \textit{Sum and Difference Identities}
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sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
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sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})
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cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})
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cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}})[/tex]
[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \\\\\\ \textit{also recall that }\qquad cos\left(\frac{\pi }{2} \right)=0\qquad sin\left(\frac{\pi }{2} \right)=1 \\\\ -------------------------------\\\\ cot\left( x-\frac{\pi }{2} \right)=-tan(x)\\\\ -------------------------------[/tex]
[tex]\bf cot\left( x-\frac{\pi }{2} \right)\implies \cfrac{cos\left( x-\frac{\pi }{2} \right)}{sin\left( x-\frac{\pi }{2} \right)}\implies \cfrac{cos(x)cos\left(\frac{\pi }{2} \right)+sin(x)sin\left(\frac{\pi }{2} \right)}{sin(x)cos\left(\frac{\pi }{2} \right)-cos(x)sin\left(\frac{\pi }{2} \right)} \\\\\\ \cfrac{cos(x)~(0)+sin(x)~(1)}{sin(x)~(0)-cos(x)~(1)}\implies \cfrac{sin(x)}{-cos(x)}\implies -tan(x)[/tex]
[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \\\\\\ \textit{also recall that }\qquad cos\left(\frac{\pi }{2} \right)=0\qquad sin\left(\frac{\pi }{2} \right)=1 \\\\ -------------------------------\\\\ cot\left( x-\frac{\pi }{2} \right)=-tan(x)\\\\ -------------------------------[/tex]
[tex]\bf cot\left( x-\frac{\pi }{2} \right)\implies \cfrac{cos\left( x-\frac{\pi }{2} \right)}{sin\left( x-\frac{\pi }{2} \right)}\implies \cfrac{cos(x)cos\left(\frac{\pi }{2} \right)+sin(x)sin\left(\frac{\pi }{2} \right)}{sin(x)cos\left(\frac{\pi }{2} \right)-cos(x)sin\left(\frac{\pi }{2} \right)} \\\\\\ \cfrac{cos(x)~(0)+sin(x)~(1)}{sin(x)~(0)-cos(x)~(1)}\implies \cfrac{sin(x)}{-cos(x)}\implies -tan(x)[/tex]
