The decreasing sine branch

(i) Express sin⁡x+3cos⁡x\sin x+\sqrt3\cos x as Rsin⁡(x+α)R\sin(x+\alpha) with exact positive R and alpha. mark

(ii) Define f by that expression on π/6≤x≤k\pi/6\le x\le k. Find the largest exact k for which f has an inverse, define the inverse including its domain, and state the inputs where ff−1(x)=f−1f(x)ff^{-1}(x)=f^{-1}f(x). marks

[2016/NYJC/Prelim/I/10a]

Total marks: 6

Solution

(i)

Auxiliary angle

Express sin⁡x+3cos⁡x\sin x+\sqrt3\cos x in the form Rsin⁡(x+α)R\sin(x+\alpha), where R>0R>0 and 0<α<π20<\alpha<\frac\pi2 are exact.

Hint

  1. Expand Rsin⁡(x+α)=Rcos⁡αsin⁡x+Rsin⁡αcos⁡xR\sin(x+\alpha)=R\cos\alpha\sin x+R\sin\alpha\cos x and match coefficients.

Answer · (i)

R=2,α=π/3R=2,\alpha=\pi/3.

(ii) restriction

One-one restrictions

f(x)=2sin⁡(x+π3)f(x)=2\sin(x+\frac\pi3) for π6≤x≤k\frac\pi6\le x\le k. Find the largest exact kk for which f−1f^{-1} exists.

Explore this idea

Hint

  1. Locate the maximum of ff at the start.

  2. The graph then decreases until the next minimum.

Answer · (ii) restriction

k=7π/6k=7\pi/6.

Explore the idea

One decreasing half-wave from maximum to minimum

The restricted graph is drawn solid; the rest of the sine wave is faint.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • F restricted
Ideas to try
What happens to one-one if kk is larger than 7π6\frac{7\pi}6?
(ii) inverse

Inverse branch selection

With this value of kk, define f−1f^{-1}, including its domain.

Explore this idea

Hint

  1. Write y=2sin⁡(x+π3)y=2\sin(x+\frac\pi3) and let u=x+π3u=x+\frac\pi3.

  2. Here uu lies in [π2,3π2][\frac\pi2,\frac{3\pi}2], not in [−π2,π2][-\frac\pi2,\frac\pi2], so use sin⁡u=sin⁡(π−u)\sin u=\sin(\pi-u).

  3. Solve for xx.

Answer · (ii) inverse

f−1(x)=2π/3−arcsin⁡(x/2)f^{-1}(x)=2\pi/3-\arcsin(x/2), −2≤x≤2-2\le x\le2.

Correct decreasing branch; printed arcsin−π/3 rule is wrong.

Explore the idea

The inverse decreases too

A decreasing function has a decreasing inverse, so the inverse here is not the increasing arcsin(x/2) − π/3.

Point on f: input xx = 2.00
Try an example

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))
(2, 0.18851)
Q = (f(x), x)
(0.18851, 2)
on the inverse
Domain of f
π6≤x≤7π6\frac\pi6\le x\le\frac{7\pi}6
Range of f = domain of f⁻¹
−2≤y≤2-2\le y\le2
Midpoint M on y = x
(1.09425, 1.09425)
  • F on its restricted domain
  • F⁻¹(x) = 2π/3 − sin⁻¹(x/2)
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

Reflecting a decreasing curve in y=xy=x gives a decreasing curve. sin⁡−1x2−π3\sin^{-1}\frac x2-\frac\pi3 is increasing, so it cannot be the inverse.

Ideas to try
  1. Check f−1(2)f^{-1}(2): it should be π6\frac\pi6, where ff has its maximum.
  2. Is the inverse increasing or decreasing?
(ii) equal composites

Equal composites

State the values of xx for which ff−1(x)=f−1f(x)ff^{-1}(x)=f^{-1}f(x).

Hint

  1. Both composites equal xx where defined.

  2. Intersect DfD_f and Df−1D_{f^{-1}}.

Answer · (ii) equal composites

π/6≤x≤2\pi/6\le x\le2.