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Answer:

#sin(arccos(x)) = sqrt(1-x^2)#

Explanation:

From Pythagoras, we have:

#sin^2 theta + cos^2 theta = 1#

If #x in [-1, 1]# and #theta = arccos(x)# then:

#theta in [0, pi]#

#sin(theta) >= 0#

Hence:

#sin(arccos(x)) = sin(theta) = sqrt(1 - cos^2 theta) = sqrt(1-x^2)#

Note we can use the non-negative square root since we have already established that #sin(arccos(x)) >= 0#

Let #a+b=x and a-b =y#

So #2a=x+y and 2b =x-y#

Now #RHS=cos2b-cos2a#

#=cos(x-y)-cos(x+y)#

#=cosxcosy+sinxsiny-cosxcosy+sinxsiny#

#=2sinxsiny#

#=2sin(a+b)sin(a-b)=LHS#

Answer:

#x=sqrt(3)/3#

Explanation:

#"arccot"(x) + 2arcsin(sqrt(3)/2) = pi#

The inverse sine function #arcsin(x)# is defined as the unique value in the interval #[-pi/2,pi/2]# such that #sin(arcsin(x)) = x#. On that interval, we have #sin(pi/3) = sqrt(3)/2# as a well known angle. Thus #arcsin(sqrt(3)/2) = pi/3#

#=> "arccot"(x) + (2pi)/3 = pi#

#=> "arccot"(x) = pi/3#

#=> cot("arccot"(x)) = cot((pi)/3)#

#=> x = cot((pi)/3)#

#=cos((pi)/3)/sin((pi)/3)#

#=(1/2)/(sqrt(3)/2)#

#=1/sqrt(3)#

#=sqrt(3)/3#

Given expression #=arcsin((x+1)/sqrt(2*(x²+1)))#
Let #x = tantheta#

So #theta=tan^-1x#

Inserting #x = tantheta# the given expression becomes

#=arcsin((tantheta+1)/sqrt(2*(tan^2theta+1)))#

#=arcsin((sintheta/costheta+1)/sqrt(2*(sec^2theta)))#

#=arcsin(1/sqrt2((sinthetasectheta+1)/(sectheta)))#

#=arcsin(1/sqrt2((sinthetacancelsectheta)/cancelsectheta+1/(sectheta)))#

#=arcsin(1/sqrt2(sintheta+costheta))#

#=arcsin(1/sqrt2sintheta+1/sqrt2costheta)#

#=arcsin(cos(pi/4)sintheta+sin(pi/4)costheta)#

#=arcsin(sin(theta+pi/4))#

#=theta+pi/4#

#=tan^-1x+pi/4#

#cos((17pi)/12)#

#cos((24pi-7pi)/12)#

#=cos(2pi-(7pi)/12)#

#=cos((7pi)/12)#

#=cos((4pi+3pi)/12)#

#=cos((4pi)/12+(3pi)/12)#

#=cos(pi/3+pi/4)#

#=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)#

#=1/2*1/sqrt2-sqrt3/2*1/sqrt2#

#=(1-sqrt3)/(2sqrt2)#

#=1/4(sqrt2-sqrt6)#

Answer:

The Inverse Sine Function is denoted by, #arc sin # or, #sin^-1# and

is defined by,

#arc sin x=theta, x in [-1,1] iff sin theta=x, theta in [-pi/2,pi/2]#

Explanation:

The Inverse Sine Function is denoted by, #arc sin # or, #sin^-1# and

is defined by,

#arc sin x=theta, x in [-1,1] iff sin theta=x, theta in [-pi/2,pi/2]#

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