The function $f(x)= ext{sin}x+ ext{cos}x-x$ has a zero near $x=1.2$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2001 - Paper 1

Using the properties of intersecting chords in a circle, we know that:

$Z AOB = 2 heta$

This is because the angle at the center (here, $AOB$) is twice the angle at circumference (${ZAB}$), which subtends the same arc AB.

Since $A$ lies on circle $C_2$ and is defined as the tangent point where line $PAT$ meets circle $C_1$, and by the properties of tangents, we know:

$ext{Z}TAB = 90^ ext{o} - ZOAP ext{; hence need to confirm that OAP = 20° because it is subtended by the same arc AB.}$

From the triangle $OAP$ and the property that angles subtended from the same segment are equal,

$ext{if } PA = OA, ext{then } BA must equal PA ext{ as triangle is isosceles.}$

Starting from the given identity:

$ext{sin}( heta + 20^ ext{o}) = ext{sin}(20^ ext{o}) ext{cos}( heta) + ext{cos}(20^ ext{o}) ext{sin}( heta)$

Using angle addition and the double angle formulas:

1. Express sin(3θ) as sin(θ+2θ).

2. Apply the sine addition formula:

   $ext{sin}(3 heta) = ext{sin}( heta) ext{cos}(2 heta) + ext{cos}( heta) ext{sin}(2 heta)$

3. Using the double angle formula:

   $ext{sin}(2 heta) = 2 ext{sin}( heta) ext{cos}( heta)$

4. Substitute this back to derive:

5. Combine like terms to reach the final identity: $ext{sin}3 heta = 3 ext{sin}( heta) - 4 ext{sin}^3( heta)$

From our derived identity:

$3 ext{sin}( heta) - 4 ext{sin}^3( heta) = 2 ext{sin}( heta)$

Rearranging leads to:

$4 ext{sin}^3( heta) - ext{sin}( heta) = 0$

Factoring out $ext{sin}( heta)$ gives:

$ext{sin}( heta)(4 ext{sin}^2( heta) - 1) = 0$

This results in:

1. $ext{sin}( heta) = 0$
2. 4 ext{sin}^2( heta) - 1 = 0 ightarrow ext{sin}^2( heta) = rac{1}{4} ightarrow ext{sin}( heta) = rac{1}{2}

Solutions for $0 ext{ ≤ } θ < 2 ext{π}$ are:

heta = 0, rac{ ext{π}}{6}, rac{5 ext{π}}{6}, ext{ and any additional solutions within range}