Lecture Notes for Math 1012 (Lecture 2) Compound Angle Formulae PDF

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Lecture Notes for Math 1012 (Lecture 2) Compound Angle Formulae

(A)

Compound Angle Formulae

1) Prove the Identity cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 .

Page | 1

(𝐾𝐿)2 = 2 − 2(cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽)

…..(1)

By Cosine Law, we have (𝐾𝐿)2 = (𝑂𝐿)2 + (𝑂𝐾)2 + 2(𝑂𝐿)(𝑂𝐾) cos(𝛼 − 𝛽)

Proof :

(𝐾𝐿)2 = 1 + 1 − 2 cos(𝛼 − 𝛽)

(𝐾𝐿)2 = 2 − 2 cos(𝛼 − 𝛽) From (1) and (2) :

….(2)

(𝐾𝐿)2 = (𝐾𝐿)2

2 − 2(cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽) = 2 − 2 cos(𝛼 − 𝛽 )

cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 (QED) 2)

Prove that

Proof :

cos(𝛼 + 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽

cos(𝛼 + 𝛽) = cos(𝛼 − (−𝛽))

= cos 𝛼 cos(−𝛽) + sin 𝛼 sin(−𝛽)

= cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽

𝐿 (𝑎, 𝑏) = (cos 𝛽 , sin 𝛽)

𝐾 (𝑥, 𝑦) = (cos 𝛼 , sin 𝛼) By the formula of Distance between two points (𝐾𝐿)2

= (cos 𝛽 − cos 𝛼

)2

+ (sin 𝛽 − sin 𝛼)

2

= (cos 𝛽)2 − 2 cos 𝛼 cos 𝛽 + (cos 𝛼)2 + (sin 𝛽)2 − 2 sin 𝛼 sin 𝛽 + (sin 𝛼)2

3)

Prove that sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵

Proof :

𝜋

As 𝑠𝑖𝑛(𝐴 – 𝐵) = cos ( 2 − (𝐴 − 𝐵)) = cos ((

𝜋 − 𝐴) + 𝐵) 2

𝜋 𝜋 = cos ( − 𝐴) cos 𝐵 − sin ( − 𝐴) sin 𝐵 2 2 = 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵

Lecture Notes for Math 1012 (Lecture 2) Compound Angle Formulae

4)

Prove that

𝑡𝑎𝑛(𝐴 − 𝐵) = sin(𝐴 – 𝐵)

Proof : 𝑡𝑎𝑛(𝐴 − 𝐵) = cos (𝐴−𝐵) =

𝑡𝑎𝑛𝐴 − 𝑡𝑎𝑛𝐵 1 + 𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵

𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵

𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵

𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 𝑡𝑎𝑛𝐴 − 𝑡𝑎𝑛𝐵 = 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 = 1 + 𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵

5)

To Sum up, we have the following are important trigonometric relationships:

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Example 1 Using the t-ratios of 30° and 45°, evaluate sin 75° Solution: 𝑠𝑖𝑛 75° = 𝑠𝑖𝑛 (45° + 30°) = 𝑠𝑖𝑛 45° 𝑐𝑜𝑠 30° + 𝑐𝑜𝑠 45° 𝑠𝑖𝑛 30° √2

√3

= ( )( 2

2

√2

1

) + ( )( ) = 2

2

√6+√2 4

Example 2 From the formula of 𝑠𝑖𝑛 (𝛼 + 𝛽) deduce the formulae of 𝑐𝑜𝑠 (𝛼 + 𝛽) and 𝑐𝑜𝑠 (𝛼 − 𝛽). Solution:

𝑐𝑜𝑠(𝐴 − 𝐵) = 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵

𝑠𝑖𝑛(𝐴 − 𝐵) = 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵 𝑡𝑎𝑛𝐴 − 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛(𝐴 − 𝐵) = 1 + 𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵

To find 𝑠𝑖𝑛(𝐴 + 𝐵), 𝑐𝑜𝑠(𝐴 + 𝐵) and 𝑡𝑎𝑛(𝐴 + 𝐵), just change the - signs in the above identities to + signs and vice-versa: 𝑠𝑖𝑛(𝐴 + 𝐵) = 𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵 𝑐𝑜𝑠(𝐴 + 𝐵) = 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 𝑡𝑎𝑛𝐴 + 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛(𝐴 + 𝐵) = 1 − 𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵

We know that, 𝑠𝑖𝑛 (𝛼 + 𝛽) = 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 + 𝑐𝑜𝑠 𝛼 𝑠𝑖𝑛 𝛽 …….. (i)

Replacing α by (90° + 𝛼) on both sides of (i) we get, sin(90° + 𝛼 + 𝛽) = sin{(90° + 𝛼 ) + 𝛽}

= 𝑠𝑖𝑛 (90° + 𝛼) 𝑐𝑜𝑠 𝛽 + 𝑐𝑜𝑠 ( 90° + 𝛼) 𝑠𝑖𝑛 𝛽, [Applying the formula of 𝑠𝑖𝑛 (𝛼 + 𝛽)]

⇒ sin{90° + (𝛼 + 𝛽)} = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽,

[since 𝑠𝑖𝑛 (90° + 𝛼) = 𝑐𝑜𝑠 𝛼 and 𝑐𝑜𝑠 (90° + 𝛼) = − 𝑠𝑖𝑛 𝛼] ⇒ 𝑐𝑜𝑠 (𝛼 + 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 − 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽 …….. (ii)

Again, replacing β by (- β) on both sides of (ii) we get, 𝑐𝑜𝑠 (𝛼 − 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 (− 𝛽) − 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 (− 𝛽) ⇒ cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽,

[since 𝑐𝑜𝑠 (− 𝛽) = 𝑐𝑜𝑠 𝛽 and 𝑠𝑖𝑛 (− 𝛽) = − 𝑠𝑖𝑛 𝛽]

Lecture Notes for Math 1012 (Lecture 2) Compound Angle Formulae

(B)

𝑠𝑖𝑛2𝐴 = 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐴

so: similarly:

cos 2𝐴 = 𝑐𝑜𝑠 2 𝐴 − 𝑠𝑖𝑛2 𝐴

Replacing 𝑐𝑜𝑠2 𝐴 by 1 − 𝑠𝑖𝑛2 𝐴 in the above formula gives: 𝑐𝑜𝑠2𝐴 = 1 − 2𝑠𝑖𝑛2 𝐴

Replacing 𝑠𝑖𝑛2 𝐴 by 1 − 𝑐𝑜𝑠2 𝐴 gives: 𝑐𝑜𝑠2𝐴 = 2𝑐𝑜𝑠2 𝐴 − 1 It can also be shown that: 𝑡𝑎𝑛2𝐴 =

2 𝑡𝑎𝑛𝐴 1 − 𝑡𝑎𝑛2 𝐴

𝑠𝑖𝑛2𝐴 = 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐴

cos 2𝐴 = 𝑐𝑜𝑠 𝐴 − 𝑠𝑖𝑛2 𝐴 = 1 − 2𝑠𝑖𝑛2 𝐴 = 2𝑐𝑜𝑠 2 𝐴 − 1 2

𝑡𝑎𝑛2𝐴 =

(C)

𝑐𝑜𝑠(𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵) = 2 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵

Double Angle Formulae

𝑠𝑖𝑛(𝐴 + 𝐵) = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 Replacing B by A in the above formula becomes: 𝑠𝑖𝑛(2𝐴) = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐴

2 𝑡𝑎𝑛𝐴 1 − 𝑡𝑎𝑛2 𝐴

Product to Sum Formulae

Sometimes it is useful to be able to write a product of trigonometric functions as a sum of simpler trigonometric functions (this might make integration easier, for example).

Now, 𝑐𝑜𝑠(𝐴 + 𝐵) = 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 and 𝑐𝑜𝑠(𝐴 − 𝐵) = 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 Adding these two:

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Subtracting one from the other: 𝑐𝑜𝑠(𝐴 − 𝐵 ) − 𝑐𝑜𝑠(𝐴 + 𝐵) = 2 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵

Similar formula can be obtained using the expansion of 𝑠𝑖𝑛(𝐴 + 𝐵). 𝑠𝑖𝑛(𝐴 + 𝐵) = 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 + 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 𝑠𝑖𝑛(𝐴 − 𝐵 ) = 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵 By adding and subtracting these two: have

And

𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) = 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵

𝑠𝑖𝑛(𝐴 + 𝐵) − 𝑠𝑖𝑛(𝐴 − 𝐵) = 2 𝑠𝑖𝑛𝐵 𝑐𝑜𝑠𝐴

2 𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 = 𝑐𝑜𝑠(𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵) 2 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 = 𝑐𝑜𝑠(𝐴 − 𝐵 ) − 𝑐𝑜𝑠(𝐴 + 𝐵) 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) 2 𝑠𝑖𝑛𝐵 𝑐𝑜𝑠𝐴 = 𝑠𝑖𝑛(𝐴 + 𝐵) − 𝑠𝑖𝑛(𝐴 − 𝐵)

(D)

Sum to Product Formulae

Now we let 𝐴 =

𝛼+𝛽 2

and 𝐵 =

𝛼−𝛽 2

,

we have 𝐴 + 𝐵 = 𝛼 and 𝐴 − 𝐵 = 𝛽 , 𝛼−𝛽 𝛼+𝛽 𝑐𝑜𝑠 2 2 𝛼−𝛽 𝛼+𝛽 𝑐𝑜𝑠 𝑠𝑖𝑛(𝛼) − 𝑠𝑖𝑛(𝛽) = 2 𝑠𝑖𝑛 2 2 𝛼+𝛽 𝛼−𝛽 𝑐𝑜𝑠 𝑐𝑜𝑠(𝛼) + 𝑐𝑜𝑠(𝛽) = 2 𝑐𝑜𝑠 2 2 𝜶+𝜷 𝜶−𝜷 𝒔𝒊𝒏 𝒄𝒐𝒔(𝜶) − 𝒄𝒐𝒔(𝜷) = −𝟐 𝒔𝒊𝒏 𝟐 𝟐

𝑠𝑖𝑛(𝛼) + 𝑠𝑖𝑛(𝛽) = 2 𝑠𝑖𝑛

Lecture Notes for Math 1012 (Lecture 2) Compound Angle Formulae

(E)

Exerceis :

Q1) Solution:

Solve 𝑠𝑖𝑛(6𝑥) + 𝑠𝑖𝑛(4𝑥) = 0 , 𝑥 ∈ [0, 2𝜋) . Using the first sum to product identity we can replace 𝑠𝑖𝑛(6𝑥) + 𝑠𝑖𝑛(4𝑥) with 2𝑠𝑖𝑛(5𝑥)𝑐𝑜𝑠(𝑥) to get 2 sin(5𝑥) cos(𝑥) = 0. We set each factor to 0 and solve the resulting simple equation. 𝑠𝑖𝑛(5𝑥) = 0 has solutions 0 and

[0,

2𝜋 5

).

𝑐𝑜𝑠(𝑥) = 0 has solutions [0, 2𝜋). The solutions are 𝑥 = 0,

Q2)

Solution:

𝜋

5

𝜋

2

and

,

𝜋 2

𝜋

5

in the interval

3𝜋

in the interval

2

and

3𝜋 2

.

Solve 𝑠𝑖𝑛(2𝑥) + 4𝑐𝑜𝑠(𝑥) = 0 for x in the interval [0,2𝜋). Replace 𝑠𝑖𝑛(2𝑥) with 2𝑠𝑖𝑛(𝑥)𝑐𝑜𝑠(𝑥) to get 2𝑠𝑖𝑛(𝑥)𝑐𝑜𝑠(𝑥) + 4𝑐𝑜𝑠(𝑥) = 0. Factor the left side to get 2𝑐𝑜𝑠(𝑥)(𝑠𝑖𝑛(𝑥) + 2) = 0. Now set each factor to 0. cos(𝑥) = 0 has solutions

𝜋

2

and

3𝜋 2

.

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𝑠𝑖𝑛(𝑥) + 2 = 0 or 𝑠𝑖𝑛(𝑥) = −2 has no solutions. The solutions are 𝑥 =

𝜋

2

,

3𝜋 2

.

Here is the graph 𝑦 = 𝑠𝑖𝑛(2𝑥) + 4𝑐𝑜𝑠(𝑥) which shows the x-intercepts in [0,2𝜋). These x-intercepts are the solutions to the equation....