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cos(atan(sqrt(3))+asin(1/3)

 May 7, 2015

Best Answer 

 #1
avatar+26396 
+10

cos( atan( sqrt(3) )+asin(1/3) ) ?

cos( arctan( 3 )+arcsin(13) )= ?

 

cos(αrad+βrad)=cos(αrad)cos(βrad)sin(αrad)sin(βrad)αrad=arctan(a)a=3βrad=arcsin(b)b=13

cos(αrad+βrad)=cos(arctan(a))cos(arcsin(b))sin(arctan(a))sin(arcsin(b))=cos(arctan(a))cos(arcsin(b))sin(arctan(a))b

 cos( arctan(a) )=±11+a2=±11+(3)2=±12

 cos( arcsin(b) )=±1b2=±1+(13)2=±83

 sin( arctan(a) )=±a1+a2=±31+(3)2=±32

cos(αrad+βrad)=(±12)(±83)(±32)(13)=16(±8±3)

cos( arctan( 3 )+arcsin(13) )=16(+8+3)=0.76007965539=16(+83)=0.18272938620=16(8+3)=0.18272938620=16(83)=0.76007965539

 

 

 May 8, 2015
 #1
avatar+26396 
+10
Best Answer

cos( atan( sqrt(3) )+asin(1/3) ) ?

cos( arctan( 3 )+arcsin(13) )= ?

 

cos(αrad+βrad)=cos(αrad)cos(βrad)sin(αrad)sin(βrad)αrad=arctan(a)a=3βrad=arcsin(b)b=13

cos(αrad+βrad)=cos(arctan(a))cos(arcsin(b))sin(arctan(a))sin(arcsin(b))=cos(arctan(a))cos(arcsin(b))sin(arctan(a))b

 cos( arctan(a) )=±11+a2=±11+(3)2=±12

 cos( arcsin(b) )=±1b2=±1+(13)2=±83

 sin( arctan(a) )=±a1+a2=±31+(3)2=±32

cos(αrad+βrad)=(±12)(±83)(±32)(13)=16(±8±3)

cos( arctan( 3 )+arcsin(13) )=16(+8+3)=0.76007965539=16(+83)=0.18272938620=16(8+3)=0.18272938620=16(83)=0.76007965539

 

 

heureka May 8, 2015
 #2
avatar+118703 
0

That looks very impressive Heureka :))

 May 8, 2015

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