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Featured 3 months ago

From Pythagoras, we have:

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

If

#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

Featured 3 months ago

Let

So

Now

Featured 3 months ago

The inverse sine function

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

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

#=1/sqrt(3)#

#=sqrt(3)/3#

Featured 3 months ago

Given expression

Let

So

**Inserting**

Featured 3 months ago

Featured 3 months ago

The Inverse Sine Function is denoted by,

is defined by,

The Inverse Sine Function is denoted by,

is defined by,

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