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Expressions of the form a cos θ + b sin θ We are unable to solve functions of the form a cos θ + b sin θ so we need to change them into an expression containing just a cos or sin. Hence we express asinθ + bcosθ in the form Rsin(θ±α) or Rcos(θ±α) whereby R and α are constants to be found. To find the values of R and α we use the compound identities, a bit of Pythagoras and some basic trigonometry. Recall the Compound Identities, these will come in useful!
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Circle the trig identities that you will need to learn for the Core 3 & 4 Exams. Remember some are given in the formula booklet so no need to learn.
Aims:• To be able to express asinθ +bcosθ in the form Rsin(θ±α) and Rcos(θ±α).• To be able to solve trig equations in the forms Rsin(θ±α) and Rcos(θ±α).• To be able to state the max and min values of trig equations in the forms Rsin(θ±α) and Rcos(θ±α).
Trigonometry Lesson 5
12 sin θ – 5 cos θ = 8
Expressions of the form a cos θ + b sin θ
We are unable to solve functions of the form a cos θ + b sin θ so we need to change them into an expression containing just a cos or sin.
Hence we express asinθ + bcosθ in the form Rsin(θ±α) or Rcos(θ±α) whereby R and α are constants to be found.
To find the values of R and α we use the compound identities, a bit of Pythagoras and some basic trigonometry.
Recall the Compound Identities, these will come in useful!
Equations of the form a cos θ + b sin θ = c
Start by writing this as an identity:
3cos + 4sin cos( )R
Using the addition formula for cos(A – B) gives:
3cos + 4sin cos cos + sin sinR R
Equating the coefficients of cos θ and sin θ :
3 = cosR 3cos =R
4 = sinR 4sin =R
Express 3 cos θ + 4 sin θ in the form R cos (θ – α).
Equations of the form a cos θ + b sin θ = c
Using these in a right-angled triangle gives:
α
R
2 2= 3 + 4 = 5R
1 43= tan
So, using these values:
3 cos θ + 4 sin θ = 5 cos (θ – 53.1°)
On w/b1. Express 2cos θ + sin θ in the form R cos (θ – α).
Equations of the form a cos θ + b sin θ = ca) Express 12 sin θ – 5 cos θ in the form R sin(θ – α).b) Solve the equation 12 sin θ – 5 cos θ = 8 in the interval 0 < θ < 360°.c) State the max value and find θ for which this occurs.
a) 12sin 5cos sin( )R Using the addition formula for sin(θ – α) gives:
12sin 5cos sin cos cos sinR R
Equating the coefficients of cos θ and sin θ :
5 = sinR 5sin =R
12 = cosR 12cos =R
Using the following right-angled triangle:
α
R2 2= 5 +12 =13R
1 512= tan
So, using these values 12 sin θ – 5 cos θ = 13 sin (θ – 22.6°)
Equations of the form a cos θ + b sin θ = c
Equations of the form a cos θ + b sin θ = c
b) Using the form found in part a) we can write the equation 12 sin θ – 5 cos θ = 8 as
13sin( 22.6 ) = 8 8
13sin( 22.6 ) =
(Using a calculator set to degrees:)
So
θ = (to 3 s.f.)
138sin6.22 1
y=8/13
b) Solve the equation 12 sin θ – 5 cos θ = 8 in the interval 0 < θ < 360°
Equations of the form a cos θ + b sin θ = c
c) The maximum value occurs when sin of the angle is _______
So 13sin(Ө - 22.6)º =
This happens when sin(Ө - 22.6)º =
so (Ө - 22.6)º =
Ө =
c) State the max value of 13 sin (θ – 22.6°) and find θ for which this occurs.
On w/ba) Express 5cos θ + 6 sin θ in the form R sin(θ + α). Where α is acute.b) State the max value and find θ for which this occurs.
1. Complete wheel puzzle in pairs2. Treasure hunt in 3 or 4’s.3. Extra practice – Ex 5B p65
To solve expressions of the form a cos θ + b sin θ write it first as:
R cos (θ – α) or R cos (θ + α) or R sin (θ – α) or R sin (θ + α)
You will be told which one to use in the exam question
Have you understood today’s lesson?
RememberExtra practice homework – Ex 5B p65 all do qu 2,3,5Challenge qu - 10