Back to: Further Mathematics SS1
BASIC TRIGONOMETRIC RATIO
The basic trigonometric ratios can be defined in terms of the sides of a right angled triangle.
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:
(a) cosӨ
(b) tanӨ
(c) secӨ
(d) cosecӨ
(e) cotӨ
First Quadrant
sinӨ = y
cosӨ= x
tanӨ = y / x
Example: Use table to evaluate (a) sin37 (b) cos75 (c) tan62
Solution
(a) sin37 = 0.6018
(b) cos75 = 0.2588
(c) tan62 = 1.881
Tan (180 – Ө) = -tanӨ
Example: Use table to evaluate (a) sin143 (b) cos 115 (c) tan 125
Solution
(a) sin143 = sin(180-143) = sin37 = 0.6018
(b) cos115 = -cos(180-115) = -cos65 = -0.4226
(c) 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
(a) sin220 = sin (180 + 40) = – sin40 = – 0.6428
(b) cos236 = cos (180 + 56) = – cos56 = – 0.5992
(c) 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
(a) sin3100 = – sin (360-310) = – sin50 = – 0.7660
(b) cos2850 = cos (360-285) = cos75 = 0.2588
(c) 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
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.
x = cosӨ …. (1)
y = 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
2) 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.
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