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dimontsho4717
dimontsho4717
20.11.2020 • 
Mathematics

I forgot to study for Finals anybody know AP Trigonometry? If x+y+z=\pix+y+z=π prove the trigonometric identity
cot{\frac{x}{2}}+cot{\frac{y}{2}}+cotg\frac{z}{2}=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}cot
x+y+z=π, so \{\frac{x}{2}}+{\frac{y}{2}}={\frac{\pi}{2}}-{\frac{z}{2}}; {\frac{z}{2}}={\frac{\pi}{2}}-({\frac{x}{2}}+{\frac{y}{2}})

). Because of \displaystyle cot\alpha={\frac{1}{tan\alpha}}cotα=
, we get \displaystyle cot\alpha={\frac{cos\alpha}{sin\alpha}}cotα=
sinα
cosα
. Then \displaystyle cot{\frac{x}{2}}+cot{\frac{y}{2}}+cot{\frac{z}{2}}={\frac{cos{\frac{x}{2}}}{sin{\frac{x}{2+{\frac{cos{\frac{y}{2}}}{sin{\frac{y}{2+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2={\frac{cos{\frac{x}{2}}sin{\frac{y}{2}}+cos{\frac{y}{2}}sin{\frac{x}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2={\frac{sin({\frac{x}{2}}+{\frac{y}{2}})}{sin{\frac{x}{2}}sin{\frac{y}{2+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2={\frac{cos{\frac{z}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2=cos{\frac{z}{2}}\cdot{\frac{sin{\frac{z}{2}}+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}sin{\frac{z}{2={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2\cdot{\frac{cos({\frac{x}{2}}+{\frac{y}{2}})+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2\cdot{\frac{cos{\frac{x}{2}}cos{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}cot
+cot
2
z
=

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