Further Trig identities

Chapter 7

After completing this chapter you should be able to:

• Use the addition formulae• Use the double angle formulae• Write expressions of the form acosϴ ± bsinϴ in

the form Rcos(ϴ ± α) and/or Rsin(ϴ ± α)• Use the factor formulae• Use all of the above to solve equations and prove

identities

The Addition Formulae

These are listed on page 107, once we’ve looked at them we’ll use them

Oh goody that’s so exciting I hear you all thinking (apart from Gamal who insists on telling us all)

Double Angle Formulae

These are on page 113, lets look and see what they’re all about.

(find the 2 other forms of this identity)

Lets work through the examples pg 113

Exercise 7B page 115 to try some

Use the double angle formulae to solve more equations and prove more identities

Prove the identity

Write down what you remember about tan2ϴ and take it from there

Show sin3A ≡ 3sinA – 4sin³A

Write down what happens if you usesin(2A + A)

examples 14 and 15 take your skills further

exercise 7C page 118

Write expressions of the form acosϴ ± bsinϴ in the form Rcos(ϴ ± α) and/or Rsin(ϴ ± α)

you can: calculate R from R² = a² + b²

α = tan-1()

these results can be deduced from example 16but can only be used as a check in Edexcel exams

to evaluate R and α you should expand the expression including R into it’s equivalent addition

formulae

e.g. Rsin(ϴ - α) = Rsinϴcosα - Rcosϴsinα Rsin(ϴ + α) = Rsinϴcosα + Rcosϴsinα

Rcos(ϴ + α) = Rcosϴcosα - Rsinϴsinα Rcos(ϴ - α) = Rcosϴcosα + Rsinϴsinα

and then equate corresponding coefficients to find R and α

example 17a) express sinx - √3cosx in the form Rsin(x - α)b) plot the graph . . . . . check the book out for this

R² = a² + b²R = √(1² + (√3)²) = 2

α = tan-1()α = tan-1()α =

this gives us sinx - √3cosx = 2sin(x - )

the quick way gives us

The longer method requires us to do this:sinx - √3cosx ≡ Rsin(x - α) ≡ Rsinxcosα – Rcosxsinαequating coefficients givesRcosα = 1 and Rsinα = √3

divide these to get tan α = √3square and add them to start finding RR²cos²α + R²sin²α = 1 + 3R²(cos²α + sin²α) = 4 R² = 4so R = 2giving us sinx - √3cosx = 2sin(x - )

express 2cosϴ + 5sinϴ in the form Rcos(ϴ - α)

the check tells us

R² = a² + b²R = √(2² + 5²) = √29

α = tan-1()α = tan-1()α = 68.2°

this gives us2cosϴ + 5sinϴ = √29cos(ϴ - 68.2)

express 2cosϴ + 5sinϴ in the form Rcos(ϴ - α)the method gives us2cosϴ + 5sinϴ ≡ Rcos(ϴ - α) ≡ Rcos ϴ cosα + Rsin ϴ sinα equating coefficients givesRcosα = 2 and Rsinα = 5

divide these to get tan α = so α = 68.2°square and add them to start finding RR²cos²α + R²sin²α = 4 + 25R²(cos²α + sin²α) = 29 R² = 29so R = √29finally producing 2cosϴ + 5sinϴ = √29cos(ϴ - 68.2)

2cosϴ + 5sinϴ = 3

√29cos(ϴ - 68.2) = 3

cos(ϴ - 68.2) = 3 ÷ √29

ϴ - 68.2 = cos -1 (3 ÷ √29)

ϴ - 68.2 = -56.1° and ϴ - 68.2 = 56.1°

ϴ = 12.1 °, 124.3 ° ( to the nearest 0.1 °)

lets try the next part and solve the equation2cosϴ + 5sinϴ = 3 for 0 < ϴ < 360°

The Factor Formulae

these formulae are derived from the addition formulae

cos

𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄≡2𝑐𝑜𝑠𝑃+𝑄2

sin𝑃 −𝑄2

cos

s

the examples show the standard manipulation of these formulae which we have looked at quite a bit now, so read them through and then try

exercise 7D