# Show that the tangent to #f(x) = sinx# at #x=1# and the tangent to #g(x) = sin^-1x# at #y=1# are equally inclined to #y=x#?

It should be intuitive that this is true of any functions that fit this criteria as the functions are inverses and so reflected in the line

In this specific case we can prove the result:

# f(x) = sinx #

Differentiating wrt

# f'(x) = cosx #

When

And for the inverse:

# g(x) = sin^-1(x) #

When

# :. sin^-1(x) = 1 #

# :. x = sin1 #

Differentiating wrt

# g'(x) = 1/sqrt(1-x^2) #

And so, when

# g'(sin1) = 1/sqrt(1-sin^2 1) #

# g'(sin1) = 1/sqrt(cos^2 1) #

# g'(sin1) = 1/cos 1 #

We can now calculate the inclinations of the tangents, Suppose:

the tangent line at

#x=1# for#f(x)# is at an inclination of#alpha#

the tangent line at#y=1# for#g(x)# is at an inclination of#beta#

the tangent line of#y=x# is at an inclination of#pi/4#

Then (by definition of the tangent and relation to the derivative), at those points;

# tan alpha = cos1 # ;# tan beta = 1/cos1 # and# tan (pi/4) = 1 #

So our aim is to show that the angle between one tangent and #y=x is the same as the other. ie that:

# beta - pi/4 = pi/4 - alpha => alpha + beta = pi/2#

Using

# tan(alpha+beta) = (tanalpha+tanbeta)/(1-tanalphatanbeta)#

# " " = (cos1 +1/cos1)/(1-cos1 1/cos1)#

# " " = (cos1 +1/cos1)/0#

# " " = oo#

# :. alpha + beta = pi/2 # QEDOr if you prefer, using

#tan(A-B)=(tanA-tanB)/(1+tanAtanB)# we have:

# tan(beta-pi/4) = (tanbeta+tan(pi/4))/(1-tanbetatan(pi/4))#

# " " = (1/cos1-1)/(1+1/cos1) #

# " " = ((1-cos1)/cos1) / ((cos1+1)/cos1) #

# " " = (1-cos1) / (cos1+1) #

# tan(pi/4-alpha) = (tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha) #

# " " = (1-cos1)/(1+cos1) #

# " " = tan(beta-pi/4) # QED

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To show that the tangents to f(x) = sinx at x=1 and g(x) = sin^-1x at y=1 are equally inclined to y=x, we need to find the slopes of both tangents and compare them to the slope of y=x.

The slope of the tangent to f(x) = sinx at x=1 can be found by taking the derivative of f(x) with respect to x and evaluating it at x=1. The derivative of sinx is cosx, so the slope of the tangent to f(x) at x=1 is cos(1).

The slope of the tangent to g(x) = sin^-1x at y=1 can be found by taking the derivative of g(x) with respect to x and evaluating it at y=1. The derivative of sin^-1x is 1/sqrt(1-x^2), so the slope of the tangent to g(x) at y=1 is 1/sqrt(1-1^2) = 1.

The slope of y=x is 1.

Since the slopes of both tangents are cos(1) and 1 respectively, and they are both equal to 1, we can conclude that the tangents to f(x) = sinx at x=1 and g(x) = sin^-1x at y=1 are equally inclined to y=x.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- Differentiate #y^2 = 4ax# w.r.t #x# (Where a is a constant)?
- What is the equation of the line that is normal to #f(x)=e^xcos^2x -xsinx # at # x=-pi/3#?
- What is the equation of the tangent line of # f(x)=(sinpix)/(cospix) # at # x=3 #?
- What is the equation of the tangent line of #y=(x^2-x)^2# at #x=-4#?
- What is the equation of the normal line of #f(x)=2x+3/x# at #x=1#?

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