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
Found 2 solutions by venugopalramana, ikleyn:
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
Answer by ikleyn(54062)  (Show Source):
You can put this solution on YOUR website! .
This question is from textbook Advanced mathematics
Evaluate:
(A) arcsin [ -(square root of 3)/2]
(B) sin [ arccos (-3/4) ]
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I looked at and read the solution to this problem in the post by @venugopalramana.
I want to say that if you, a reader, answer this way on an exam or in an interview, an unsatisfactory score is guaranteed for you.
Below is my correct solution to both questions.
(A) arcsin, by its definition, is an angle between  and  .
There is only one angle in this interval with sine equal to  .
This angle is  , or, which is the same, -60°.
ANSWER.  = = -60°.
Solved.
(B)  , by the definition, is an angle between 0 and  , whose cosine equals  .
Since the cosine value is negative, this angle is in QII, the second quadrant.
In the second quadrant, the sine is positive. So, using the Pythagorean identity, we write
 =  =  =  =  .
ANSWER. = .
Solved.
Both questions answered correctly.
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After reading my post (s), please don't think I have something against the tutor @venugopalramana.
My only concern is that visitors are taught the right way on this forum and that readers get the right solutions here.
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