SOLUTION: Evaluate:
A.) Arcsin [ -(square root of 3)/2]
B.) sin [ Arccos (-3/4) ]
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Question 66819This question is from textbook Advanced mathematics
: Evaluate:
A.) Arcsin [ -(square root of 3)/2]
B.) sin [ Arccos (-3/4) ]
This question is from textbook Advanced mathematics
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Evaluate:
A.) Arcsin [ -(square root of 3)/2]=X SAY
SIN(X)=-SQRT(3)/2=-SIN(60)=SIN(180+60)=SIN(240)...OR...
...SIN(360-60)=SIN(300)
HENCE X=240 OR 300....GENERAL FORMULA IS
240,240+360,240+2*360......240+N*360.OR
300,300+360,300+2*360....300+N*360
B.) sin [ Arccos (-3/4) ] =
LET ARCCOS(-3/4)=Y.....WE HAVE TO FIND SIN(Y)=?
COS(Y)=-3/4..THAT IS Y IS IN II OR III QUADRANTS
SIN(Y)=SQRT[1-COS^2(Y)]=SQRT[1-9/16]=SQRT(7/16)=[SQRT(7)]/4
IF Y IS IN II QUADRANT THEN ANSWER IS [SQRT(7)]/4
IF Y IS IN III QUADRANT THEN ANSWER IS -[SQRT(7)]/4
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