Reciprocal and inverse circular functions

Two different meanings

A reciprocal changes a nonzero output. Cosecant is one divided by sine, secant is one divided by cosine, and cotangent is cosine divided by sine. An inverse undoes a one-to-one function: arcsin, arccos and arctan return angles. In this lesson inverse sine means arcsin, never cosecant. Angles are in radians

What does sin⁻¹(1/2) mean here?

Undefined and zero are different

At π, cosecant and cotangent are undefined because sinπ=0, while secπ=−1. At π/2, cotangent is0 because its numerator is0 and denominator is1; secant is undefined. Use the denominator restriction before dividing

Find cot(π/2)

Cosecant keeps sine’s period and loses its zeros

The domain excludes x=nπ for all integers n. Its period is2π and range is (−∞,−1]∪[1,∞). At sine’s extrema, the reciprocal has the same value±1. Near each excluded zero, magnitude grows without bound

Which cosecant values are possible?

Secant poles occur at cosine’s zeros

Secant excludes x=π/2+nπ, has period2π and the same range as cosecant. It is not obtained by swapping coordinates of cosine; it reciprocates cosine’s outputs

Where are secant’s vertical asymptotes?

Cotangent is not defined wherever tangent is defined

Define cotangent as cosine divided by sine. Its domain excludes nπ, its period isπ and its range is all real numbers. Each branch decreases from positive infinity to negative infinity. The reciprocal-of-tangent shorthand only applies where tangent exists and is nonzero

Does cotangent exist at π/2 even though tangent does not?

Transform the argument before locating poles

For \(y=-2\sec(3(x-\pi/6))+1\), the period is \(2\pi/3\). Solve

to get poles

Its range is

The minus sign exchanges the branches; the magnitude 2 doubles the gap width

Give the coefficient of π in the period of y=−2 sec(3(x−π/6))+1.

Give the positive coefficient of π in the pole spacing of y=−2 sec(3(x−π/6))+1.

Find the upper edge of the excluded range gap of y=−2 sec(3(x−π/6))+1.

Restrict before inverting

Sine on [−π/2,π/2], cosine on [0,π], and tangent on (−π/2,π/2) are one-to-one. Reflect these restricted graphs in y=x. Their ranges become inverse domains; their restricted input intervals become inverse ranges

Which is the domain of arcsin?

Which is the range of arccos?

Are the horizontal lines y=±π/2 attained by arctan?

An exact inverse value needs its principal interval

The sign of the input does not put every inverse output in the same quadrant. In particular, arccos returns an angle in \([0,\pi]\)

Give the coefficient ofπ in arccos(−1/2)

Cancellation works only on the correct domain

The reverse composition returns the original angle only in the principal interval

For the following angle, the sine is one half, so the principal output differs

Give the coefficient ofπ in arcsin(sin(7π/6))

Inverse transformations retain the inner input bound

For the transformed function

require

Thus the domain is

Multiplying the principal sine range by negative two and adding pi gives

The graph decreases; order the endpoints after reflection

Give the left domain endpoint of y=−2 arcsin(3(x−1/2))+π

Give the coefficient of π at the upper range endpoint of y=−2 arcsin(3(x−1/2))+π

Cosine inversion reverses direction before transformations

Arccos is decreasing. For y=2 arccos(−2(x+1)), the domain is[−3/2,−1/2] and range[0,2π]. The negative inner coefficient reverses the original decrease, so the transformed graph increases

Arctangent transformations have no finite domain boundary

For

the domain is all real numbers. The centre is

Its open range is

The boundary values are horizontal asymptotes and are excluded; the function decreases

Give the coefficient of π in the lower horizontal asymptote of y=−3 arctan(2(x−1))−π

Check the inverse range before applying a circular function

Solve

Isolate the inverse value

This lies in the principal range, so

By contrast, the following equation has no solution because its right side is outside the principal arcsine range

Taking sine alone would suggest the false candidate

Give x in 2 arcsin(3(x−1/2))+π=4π/3.

Solve reciprocal levels on a stated interval

Solve

First obtain

The inner angles in the interval \([0,2\pi)\) are

Therefore

Pole exclusions and the right-open interval prevent invalid or duplicate endpoints

For −2 sec(2(x−π/4))+1=−3 on x∈[π/4,5π/4), give the coefficient of π in the smaller solution

Construct and explain the reciprocal graphs, ungraded

Sketch csc,sec and cot over two periods. Label exact zeros where present, range edges, periods and every vertical asymptote. Explain why cot(π/2)=0 is consistent with the failure of1/tan(π/2). This drawing and justification are ungraded

Model: denominators determine poles

Cosecant and cotangent exclude nπ; secant excludesπ/2+nπ. Cosecant/secant have period2π and range outside the open interval(−1,1), including±1; neither has zeros. Cotangent has periodπ, all-real range and zerosπ/2+nπ. Draw separate branches continuing toward the dashed poles. The ratio cos/sin defines cotangent atπ/2; the expression1/tan has a smaller domain, so the algebraic shorthand cannot extend there

Justify the principal inverse, ungraded

Explain why reflection of the full sine graph is not an inverse function, why the chosen restriction fixes this, and why arcsin(sin θ) can differ fromθ. Give a counterexample and state both compositions with their exact domains. This proof is ungraded

Model: distinguish outputs from restricted inputs

Full sine repeats outputs, so reflecting it produces more than one output for some inverse inputs. Restricted sine is one-to-one from[−π/2,π/2] onto[−1,1], so its reflected relation is a function. Thus sin(arcsin t)=t on[−1,1], and arcsin(sin θ)=θ on[−π/2,π/2]. At5π/6 the latter composition returnsπ/6. Equal sine values alone do not select the unrestricted original angle

Check transformed graphs with technology, ungraded

Use radian mode to compare the transformed arcsin and arctan examples with your analytical domains,ranges,directions and asymptotes. Explain how a finite screen window can hide an excluded endpoint or a disconnected branch. This investigation is ungraded

Model: verify exact structure before trusting pixels

The transformed arcsin has included endpoint pairs(1/6,2π),(5/6,0) and decreases through(1/2,π). The transformed arctan decreases through(1,−π), approachingπ/2 on the left and−5π/2 on the right without attaining either. Use sample values as error checks and the principal ranges as the justification. A short plotted gap or an apparent contact with an asymptote is a display approximation, not a domain decision