Trigonometry Revision

θ

HypotenuseOpposite

Adjacent

O

S H

A

C H

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T A

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OppSin

Hyp

AdjCos

Adj

OppTan

Trig Graphs1 2 – 1 – 2

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y = Sin (x)

For y = a sin (bx)

a = ½ the total height of the graph

b = The number of times the graph repeats over 3600

1 2 3 4 5 – 1 – 2 – 3 – 4 – 5

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Here the total height of the graph is 8. a =4

The graph repeats twice over 3600. b =2

y = a sin (bx)

Hence the equation of the graph is y = 4 sin 2x

y = a cos (bx)

Here the total height of the graph is 10. a = 5

The graph repeats 3 times over 3600. b = 3

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

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Hence the equation of the graph is y = 5 cos 3x

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

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y = a sin (bx) + c

Here the total height of the graph is 4. a = 2

The graph repeats twice over 3600. b =2

The centre line has been moved up 3 places. c = 3

Hence the equation of the graph is y = 2 sin 2x +3

Find the equation of the graphs shown below.

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

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1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

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y = 2 cos(3x) -4

y = 5 sin (½ x)

Sketch the graphs of :

(i) y = 2 sin(3x)(ii) y = cos(x) -3

1 2 3 4 – 1 – 2 – 3 – 4

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Trigonometric Equations

AS

T C

θ

Once the acute solution to a trig equation is found the other solution is found using the diagram above.

180 - θ

180 + θ 360 - θ

For example: (i) Solve sin θ = ½

θ = sin-1(½)Since sin θ is + θ lies in the 1st

And 2nd quadrant

θ = 300 and 1800 - 300

θ = 300 and 1500

Example 2: Solve cos θ = -½

AS

T C

Since cos θ is negative our solutions will lie in:

the 2nd and 3rd Quadrant.

To solve all trig equations find the acute value of θ first. Then use it to find the actual solutions.

It may be that as in the previous problem, the acute value of θ is one of the actual solutions.

Acute value of θ = cos-1 (½) = 600

Actual value of θ = 1800 – 600 and 1800 + 600

θ = 1200 and 2400

(1) Solve for

00 360

0

00 360

0

(i) Cos θ = 0.7 (ii) 3 sin θ + 1 = 0

(i) AS

T C

AS

T C

Since Cos θ is positive the solutions lies in the 1st and 4th quadrant.

Acute value of θ = cos-1 0.7

= 45.60

Actual values of θ are

45.60 and 3600 – 45.60

θ = 45.60 and 314.40

3 sin θ + 1 = 0 sin θ = -1/3

Since sin θ is negative the solutions lies in the 3rd and 4th quadrant.

Acute value of θ = sin-1 (1/3)

= 19.50

Actual values of θ are1800 + 19.50 and 3600 – 19.50

θ = 199.50 and 340.50

(ii)