Week 8 – SS1 Second Term Further Mathematics Notes

WEEK EIGHT                                             DATE…………… TOPIC : TRIGONOMETRIC RATIO CONTENT

• Basic trigonometric Ratio
  
• Ratio of General Angle
  
• Trigonometric Identities
  

BASIC TRIGONOMETRIC RATIO

The basic trigonometric ratios can be defined in terms of the sides of a right angled triangle.
Q
r
p

 
 R q P
▲PQR in the figure above is a right angle triangle with QPR = Ө and PRQ = 90˚ We define the three basic ratios as follows:

      Cosine of angle Ө = PR
q
              PQ = r
    
      Sine of angle Ө =QR
p
PQ = r
    
      Tangent of angle Ө = QR p
              PR = r

 The cosine of angle Ө, sine of angle Ө and the tangent of angle Ө will be abbreviated as cosӨ, sinӨ and tanӨ respectively.

 Thus: cosӨ = q ,sinӨ = p ,tanӨ = p r r q
Also,
    sinӨ = p/r = p = tanӨ

cosӨ q/r q

 tanӨ = sinӨ
cosӨ

 Reciprocals of Basic Ratios
We define the reciprocals of the three basic ratios as:
Secant of angle Ө = PQ/PR = r/q = 1 / cosine of angle Ө.
Cosecant of angle Ө = PQ / QR = r / q = 1 / sine of angle Ө
Cotangent of angle Ө = PR / QR = q / p = 1 / tangent of angle Ө
The secant of angle Ө, the cosecant of angle Ө and the cotangent of angle Ө are abbreviated secӨ, cosecӨ and cotӨ respectively.
    SecӨ = r / q = 1 / cosӨ
    CosecӨ = r / p = 1 / sinӨ
    CotӨ = q / p = 1 / tanӨ = cosӨ / sinӨ

Example 1

Given that sinӨ = 5 / 13 and Ө is acute, find:

1. cosӨ
2. tanӨ
3. secӨ
4. cosecӨ
5. cotӨ

 Solution
Q

 
 13
5

 
 
  R 5 P
Use Pythagoras theorem to find PR
    PQ² = PR² + QR²
    13² =PR² + 5²     PR² = 13² – 5²
     = 169 – 25
     = 144
    PR = 12

 Thus, q = 12, r = 13, p = 5.

1. cosӨ = q / r = 12 /13
2. tanӨ = p / q = 5 / 12
3. secӨ = r / q = 13 / 12
4. cosecӨ = r / p = 13 / 5
5. cotӨ = q / p = 12 / 5

 

cosӨ= x
tanӨ = y / x

 Example: Use table to evaluate (a) sin37 (b) cos75 (c) tan62
Solution
(a)     sin37 = 0.6018

Cos (180 – Ө) = -cosӨ
Tan (180 – Ө) = -tanӨ

 Example: Use table to evaluate (a) sin143 (b) cos 115 (c) tan 125 Solution

1. sin143 = sin(180-143) = sin37 = 0.6018
2. cos115 = -cos(180-115) = -cos65 = -0.4226
3. tan125 = -tan(180-125) = -tan55 = -1.428

 
Third Quadrant
Sin (180 + Ө) = – sinӨ
Cos (180 + Ө) = – cosӨ
Tan (180 + Ө) = tanӨ
Example: Use table to evaluate (a) sin220 (b) cos236 (c) tan242 Solution

1. sin220 = sin (180 + 40) = – sin40 = – 0.6428
2. cos236 = cos (180 + 56) = – cos56 = – 0.5992
3. tan242 = tan (180 + 62) = tan 62 = 1.881

 Fourth Quadrant
Sin (360 – Ө) = – sinӨ
Cos (360 – Ө) = cosӨ
Tan (360 – Ө) = – tanӨ

 
 Example: Use table to evaluate (a) sin310⁰ (b) cos285⁰ (c) tan334⁰ Solution

1. sin310⁰ = – sin (360-310) = – sin50 = – 0.7660
2. cos285⁰ = cos (360-285) = cos75 = 0.2588
3. tan334⁰ = – tan (360-334) = – tan26 = – 0.4877

  Note that:

1. In the first quadrant, all the ratios are positive.
2. In the second quadrant, only sine ratio is positive, while the rest are negative.
3. In the third quadrant, only tangent ratio is positive, while the rest are negative.
4. In the fourth quadrant, only cosine ratio is positive, while the rest are negative.

Evaluation

1) Use tables to evaluate the following (a) Sin 162⁰ (b) Cos 283⁰ (c) Tan 325⁰ (d) Cos( – 75) (e)Tan (-100) (f) Sin ( -223) 2) Use tables to find the values ϕ between 0⁰ and 360⁰ which satisfy each of the following.
(a) Sin ϕ = 0.4396 (b) Tan ϕ = – 2.4398 (c) Cos ϕ = 0.8427

TRIGONOMETRIC IDENTITY

Pythagoras theorem. Y
P 1 y x
O x N

 
 The figure above shows a unit circle. ▲OPN is a right angled with OP = 1, ON= x and PN = y, PON = Ө. From the definition of trigonometric ratios.

1. = cosӨ      …. (1)
2. = sinӨ      …..(2)

From (1) x² = cos²Ө …… (3)
From (2) y² = sin²Ө …… (4) Adding equations (3) and (4) x² + y² = cos²Ө + sin²Ө …..(5)
Since ▲OPN is a right angled triangle
    ON² + NP² = OP²
    x² + y² = 1 …… (6) Equating equations (5) and (6)
    Cos²Ө + sin²Ө = 1 …… (7)
Dividing both sides of (7) by cos²Ө
    Cos²Ө / cos²Ө + sin²Ө / cos²Ө = 1 / cos²Ө
    1 + tan²Ө = sec²Ө …..(8)
Dividing (7) through by sin²Ө
    Cot²Ө + 1 = cosec²
    1 + cot²Ө = cosec²Ө …..(9)

Evaluation 1

1. Prove that (1 – Sinϕ)(1 + Sin ϕ) = Cot²ϕ

Sin² ϕ

1. Show that (Sec ϕ – Tan ϕ)(Sec ϕ + Tan ϕ)= 1

Evaluation 2

Find the values of Ѳ lying between 0 and 360 for each of the following
1)cos Ѳ = 0.2874 2)sin Ѳ = 0.9361
3)cos Ѳ =-0.8271
4)tan Ѳ =-2.106

GRAPH OF SINE AND COSINE FOR ANGLES

In the figure below, a circle has been drawn on a Cartesian plane so that its radius, OP, is of length 1unit. Such a circle is called unit circle.

 
 
 
 
 
 
 
 
 
 
 
 
 The angle Ѳ that OP makes with Ox changes according to the position of P on the circumference of the unit circle. Since P is the point (x,y) and /OP/ = 1 unit,
Sin Ѳ = y/1 = y
Cos Ѳ = x/1 = x Hence the values of x and y give a measure of cos Ѳ and sin Ѳ respectively.
If the values of Ѳ are taken from the unit circle, they can used to draw the graph of sin Ѳ. This is done by plotting values of y against corresponding values of Ѳ as in figure below.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 In the figure above, the vertical dotted lines gives the values of sin Ѳ corresponding to Ѳ = 30, 60,90,……., 360. To draw the graph of cos Ѳ , use corresponding values of x and Ѳ. This gives another wave-shaped curve, the graph of cos Ѳ as in figure below.

 
 
 
 
 
 
 
 
 
 
 
 
 
 As Ѳ increases beyond 360, both curves begin to repeat themselves as in figures below. Notice the following:
1)All values of sin Ѳ and cos Ѳ lie between +1 and -1.
2)The sine and cosine curves have the same shapes but different starting points.
3)Each curve is symmetrical about its peak(high point) and trough(low point). This means that for any value of sin Ѳ there are usually two angles between 0 and 360; likewise cos Ѳ. The only exceptions to this are at the quarter turns, where sinѲ and cosѲ have the values given in table below;

 Examples
1) Reffering to graph on page 211 of NGM Book 1, (a)Find the value of sin 252, b)solve the equation 5 sin Ѳ = 4

Solution

a)On the Ѳ axis, each small square represents 6. From construction a) on the graph: Sin 252 = -0.95 b) If 5 sin Ѳ = 4 then sin Ѳ = 4/5 = 0.8
From construction (b) on the gragh: when sin Ѳ = 0.8, Ѳ = 54 or 126

Graph of tan Ѳ

Values can be taken from a unit circle to draw a tangent curve. In the figure below, a tangent is drawn to the unit circle OX. A typical radius is drawn and extended to meet the tangent at T. the y – coordinates of T gives a measure of tan Ѳ, where Ѳ is the angle that the radius makes with OX.
Note that tan Ѳ is not defined when Ѳ =90⁰ and 270⁰.

Ratio of special Angles (45⁰, 30⁰ and 60⁰) A. Tan, Sin and Cos of 45⁰

Considering the diagram below;

 
 
 
 
 
 
 
 
 ∆ABC is right –angled triangle at B and /AB/ = /BC/ = 1 unit
/AC/² = 1² + 1² = 2 (by Pythagoras’ theorem)
/AC/ =√2
Thus, tan45⁰ = 1

B. Tan, Sin and Cos of 30⁰ and 60⁰

Considering the diagram below;

 
 
 
 
 
 
 
 
 
 
 ABC is an equilateral triangle with sides of 2 units in length. Line AD is an altitude where /BD/ = /DC/ = 1 unit. In ∆ABD, /AB/² = /AD/² + /BD/² (by Pythagoras’ theorem)
2² = /AD/² + 1²
/AD/² = 3
/AD/ = √3 units
Since, <B = 60⁰
Thus, Tan 60⁰ = √3     

Cos 60⁰ = ½ = 0.5⁰
Also, <BAD = 30⁰
Tan 30⁰ = 1/√3     
Sin 30⁰ = ½ = 0.5⁰

Example: Write the value of each the following in surd form;

1. sin135⁰
2. tan330⁰
3. cos240⁰

Solution

1. sin135⁰ = sin(180 -135) = sin 45 =
2. tan330⁰ = -tan(360-330) = tan30 = 1/√3
3. cos240⁰ = cos(240 -180) = cos60 = – 1/2

 Evaluation:
1)Using the same graph used in the above example, find the values of the following a)sin 24 b) sin 294
2)Use the same graph to find the angles whose sines are as follows: a) 0.65 b)-0.15

GENERAL EVALUATION

1. Use tables to evaluate each of the following (i) sin310 (ii) tan242 (iii) cos(-243) (iv) sin(-260) (iv) tan(-255)
2. Use tables to find the values of Ѳ between 0⁰ and 360⁰ which satisfy each of the following (i) cos Ѳ = -0.4540 (ii) tan Ѳ= 1.176 (iii) sin Ѳ = -0.9336
3. Using the same axis, a scale of 1cm to represent 30⁰ on the Ѳ-axis and 2cm to represent 1 unit on the y-axis, draw the graph of the following relations (i) y = sin Ѳ (ii) sin Ѳ/2     
4. Simplify

3- √2

3√5 + √2
READING ASSIGNMENT
NGM BK 1 PG 187 – 195; Ex 17c nos 3 and 6

WEEKEND ASSIGNMENT

Given that sin Ө = 4/5 and Ө is acute

1. find cos Ө (a) 5/3 (b) 3/5 (c) 4/3 (d) 4/5
2. find tan Ө (a) 4/5 (b) 3/5 (c) 5/4 (d) ¾
3. find cosec Ө (a) 4/5 (b) 3/5 (c) 5/4 (d) ¾
4. find sec Ө (a) 5/3 (b) 5/4 (c) ¾ (d) 5/2
5. find cot Ө (a) 3/5 (b) 4/5 (c) 5/4 (d) 5/3

THEORY

1a.) Prove that 1 + 1 = 2 cosec² Ө
    1 + cos Ө     1 – cos Ө
b.) Given that sin Ө = 5/ 13 and Ө is acute, find (i) cos Ө (ii) tan Ө (iii) sec Ө (iv) cosec Ө (v) cot Ө
2a) Copy and complete the table below, giving corresponding values of Ѳ from 0^o to 360^o Ѳ 0 30 60 90 120 150 180 210 240 270 300 330 360 Cos Ѳ 1 0.87 0.5 0 – 0.5
b)Hence draw the graph of cos Ѳ, using 2cm to 0.5 on y-axis and 1cm to 30^o x-axis

 bi) Construct a table for y = cosx – 3sinx for values of x from 0⁰ to 180⁰ at intervals of 20⁰. ii) Using a scale of 2cm to 20⁰ on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y= cosx -3sinx. iii) Use your grah to find the value(s) of x correct to the nearest degree for which (i) 3tanx = 1(ii) 2 + cosx = 3sinx.