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Trigonometric Identities


EvoN1020v

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[math]\cos{\tfrac{\pi}{12}}=\cos{(\tfrac{\pi}{3}-\tfrac{\pi}{4})}[/math]

 

[math]\cos{\tfrac{-\pi}{12}}=\cos{(\tfrac{\pi}{4}-\tfrac{\pi}{3})}[/math]

 

 

Take the double angle form on either of them. :)

 

[math]\cos{\tfrac{\pi}{3}-\tfrac{\pi}{4}}=\cos{\tfrac{\pi}{3}} \cdot \cos{\tfrac{\pi}{4}}+\sin{\tfrac{\pi}{3}} \cdot \sin{\tfrac{\pi}{3}}[/math]

 

[math]\cos{\tfrac{\pi}{4}-\tfrac{\pi}{3}}=\cos{\tfrac{\pi}{4}} \cdot \cos{\tfrac{\pi}{3}}+\sin{\tfrac{\pi}{3}} \cdot \sin{\tfrac{\pi}{4}}[/math]

 

You'll note that either way, you end up with the same (because it doesn't matter the order in which you multiply, a*b=b*a)

 

general case...

 

(where a-b=c)

 

[math]\cos{c}=\cos(a-b)[/math]

[math]\cos{-c}=\cos(b-a)[/math]

 

[math]\cos(a-b)=\cos{a} \cdot \cos{b} + \sin{a} \cdot \sin{b}[/math]

[math]\cos(b-a)=\cos{b} \cdot \cos{a} + \sin{b} \cdot \sin{a}[/math]

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