Any Help Will be Appreciated. Thanks.
1. a f(-x)= sin ( cos^-1 (-x) ) = sin ( pi- cos^-1 x ) = sin ( cos^ -1 (x) ) { sin ( 180 -x) = sin(x) as we are in 2nd quadrant } = f(x). Thus even function.
b. range cos ^ (-1) x : 0 <= y <= pi, now looking at the sin graph, it is above the x axis between x = 0 and x=pi. which means were are putting values between 0 and pi INTO the sin function, the sin function will only GIVE out values >=0
c. y= sin ( cos ^ -1 (x) )
y^2 = [ sin ( cos ^ -1 (x) ) ] ^2, let u = cos ^ (-1) x , cos(u) =x
so we have y^2 = [sinu]^2, by the substitution
y^2 = 1-(cosu)^2= 1- (x^2) [ cosu = x], take sqrt both sides, we get + or -, but as we know from part b that it is >=0, we take the +
d.
just a semicircle, radius 1 , centred at the origin,its the top half of the circle x^2 +y^2 =1 .
e. similar to d