Mathematics Extension 1 Notes

Chapter 6 – Trigonometry 6.1: Trigonometric Ratios sinθ =

1 -----cosecθ

cosθ =

1 -----secθ

cotθ =

cosθ -----sinθ

tanθ =

1 -----cotθ

tanθ =

sinθ -----cosθ

θ

6.2: Fundamental Trigonometric Identities sin²θ + cos²θ = 1 - Substitute sinθ / cosθ into unit circle equation - Dividing through by sin²θ sin²θ cos²θ 1 ------- + ------- = ------sin²θ sin²θ sin²θ



1 + cot²θ = cosec²θ



1 + tan²θ = sec²θ

When solving trigonometric proofs, always work from one side to the other until it equals the other side

- Dividing through by cos²θ sin²θ cos²θ 1 ------- + ------- = ------cos²θ cos²θ cos²θ 6.3: The Cosine Rule b² + c² - a² cos A = -------------2bc a² = b² + c² - (2bc.cos A)

Given 3 sides, asked for an angle Given 2 sides / 1 angle, asked for a side

? b

c b a

? A c

6.4: The Sine Rule

6.5: Sum and Difference of Angles (Extension 1) ~ sin (A + B) = sinA . cosB + cosA . sinB ~ sin (A – B) = sinA . cosB – cosA . sinB e.g: find the exact value of sin75º sin75º

= sin (45º + 30º) = (sin45º . cos30º) + (sin30º . sin45º) =

1 x √3 1x1 -------- + -------√2 x 2 √2 x 2



1 + √3 = -------2√2

~ cos (A + B) = cosA . cosB - sinA . sinB ~ cos (A – B) = cosA . cosB + sinA . sinB e.g: find the exact value of cos105º cos105º

= cos (60º + 45º) = (cos60º . cos45º) + (sin60º . sin45º) =

1x1 √3 x √2 -------- + -------√2 x 2 2x1



1 + √6 = --------2 + 2√2



tanθ + 1 = ---------- - 1 1 – tanθ

~ tan (A + B) = tanA + tanB ------------------1 – tanA . tanB ~ tan (A - B) = tanA - tanB ------------------1 + tanA . tanB e.g: expand and simplify tan (θ + 45) – 1 tan (θ + 45) – 1

=

=

tanθ + tan45º -------------------- - 1 1 - tanθ + tan45º

tanθ + 1 – (1 – tanθ) ----------------------- 1 – tanθ

2tanθ = ---------1 – tanθ

6.6: Ratios of Double Angles (Extension 1) Using the sum and difference of angles:



sin (A + B) = sinA . cosB + sinB . cosA - now, let B = A :. sin (A + A) = sinA . cosA + cosA . sinA sin2A = 2 (sinA . cosA)



cos (A + B) = cosA . cosB - sinA . sinB - now, let B = A :. cos (A + A) = cosA . cosA - sinA . sinA cos2A = cos²A - sin²A also, by substituting (1 - sin²A) for cos²A :. cos2A = 1 - sin²A - sin²A = 1 – 2sin²A Also, by substituting (1 - cos²A) for sin²A :. cos2A = cos²A – (1 - cos²A) = cos²A + cos²A – 1 = 2cos²A – 1



tan (A + B) = tanA + tanB ------------------1 – tanA . tanB

tanA ≠ ± 1

- now, let B = A :. tan (A + A) =

tanA + tanA -----------------1 – tanA . tanA

tan2A = 2tanA ----------1 - tan²A

X

6.7: The t = tan —(2 Let

X

t = tan —-( 2

) formula (Half-Angle Formula) )

Using the Double Angle Formula 

sin2θ = 2 (sinθ . cosθ) Divide through by 2 θ :. sinθ = 2 (sin—2

θ cos—2

)

θ sub in “t” for—2

1 1 = 2 . ---------- . ---------√(1 + t²) √(1 + t²)



2t :. sinθ = -------1 + t²

cos2θ = cos²A - sin²A Divide through by 2 θ —:. cosθ = cos²2

θ —- sin²2

θ —sub in “t” for2

1 t² = ---------- - ---------√(1 + t²) √(1 + t²)



1 - t² :. cosθ = -------1 + t²

2tanθ tan2θ = -----------1 - tan²θ Divide through by 2 θ —-

2tan 2 :. tanθ = ----------------sub in “t” for θ 1 - tan²—2

θ —-

2

2t :. tanθ = -------1 - t²

6.8: Solving using Transformation (Subsidiary / Auxiliary Angles) (Extension 1) b tan  = --a

similarly:

b sin  = ----------√(a² + b²) a cos  = ----------√(a² + b²)

b

√(a² + b²)

Let √(a² + b²) = R :.



R . sin  = b a

R . cos  = a Show that: R . sin (θ + ) = (a . sin θ) + (b . cosθ) LHS:

 Proof

R . sin (θ + ) = R . (sinθ . cos) + R . (cos . sinθ) = a . sinθ + b . cosθ

= RHS

:. R . sin (θ + ) = (a . sinθ) + (b . cosθ) Where √(a² + b²) = R, and  is in the first quadrant Furthermore: ~

R . sin (θ - ) = (a . sinθ) - (b . cosθ)

~

R . cos (θ + ) = (a . cosθ) - (b . sinθ)

~

R . cos (θ - ) = (a . cosθ) + (b . sinθ)

e.g: solve √3 . sinθ + cosθ = 1 LHS:

√3 . sinθ + cosθ

where 0º ≤ θ ≤ 360º note:

= 2sin (θ + 30) = 1 sin (θ + 30)

R = √(a² + b²)

= √(3 + 1)

a = √3

=2

= 0.5 *

b=1

:. θ = 0º, 120º, 360º note: 30º * find where (θ + 30) = 0.5 i.e: quadrants 1 and 2 (and 360º)

6.9: Products as Sums and Differences (Extension 1)

b tan  = --a



1 = --√3



=

Using the addition theorems: 1. sin (A + B) = (sinA . cosB) + (sinB . cosA) 2. sin (A – B) = (sinA . cosB) – (sinB . cosA) 3. cos (A + B) = (cosA . cosB) – (sinA . sinB) 4. cos (A – B) = (cosA . cosB) + (sinA . sinB) ~ Add 1 + 2 :. 2 (sinA . cosB) = sin (A + B) + sin (A – B)

5.

= sin (sum) + sin (difference) ~ Subtract 2 – 1 :. 2 (cosA . sinB) = sin (A + B) – sin (A – B)

6

= sin (sum) – sin (difference) ~ Add 3 + 4 :. 2 (cosA . cosB) = cos (A + B) + cos (A – B) = cos (sum) + cos (difference)

7.

~ Subtract 4 – 3 :. 2 (sinA . sinB) = cos (A – B) – cos (A + B) = cos (difference) – cos (sum)

6.10: Sums or Differences as Products (Extension 2)

8.

5.

.

Using the product theorems: 5. 2 (sinA . cosB) = sin (A + B) + sin (A – B) 6. 2 (cosA . sinB) = sin (A + B) – sin (A – B) 7. 2 (cosA . cosB) = cos (A + B) + cos (A – B) 8. 2 (sinA . sinB) = cos (A – B) – cos (A + B) ~ let U = (A + B);V = (A – B) U+V :. A = -------2

U-V B = -------2

Substitute values of U and V into 5  8 U+V U-V 9. sinU + sinV = 2sin -------- cos -------2 2 U–V U+V 10. sinU – sinV = 2sin -------- cos -------2 2 U+V U-V 11. cosU + cosV = 2cos -------- cos -------2 2

12.

a)

U+V U-V cosV – cosU = 2sin -------- sin -------2 2

b)

U+V U-V cosU – cosV = -2sin -------- sin -------2 2

e.g: express sin6x – sin4x in terms of products 6x – 4x 6x + 4x sin6x – sin4x = 2sin ( ---------- ) . cos ( ---------- ) 2 2 2x 10x = 2sin ----- . cos ----2 2



= 2sinx . cos5x