Application of Surds to Trigonometric Ratios

Welcome to class!

In today’s class, we will be talking about the application of surds to trigonometric ratios. Enjoy the class!

Application of surds to trigonometric ratios

Apart from solving quadratics, surds also arise in trigonometry.

The angles 30°, 45°, 60° have the following trigonometric ratios.

Triangle ABC is equilateral. AD is the line interval from A to the midpoint of BD. Triangles ABD and ACD are congruent (SSS).

$A n g l e s A B D = 60 °, B A D = 30 ° a n d P y t h a g o r a s ’ t h e o r e m g i v e s t h a t A D = \sqrt 3$

Hence,

sin 30° = cos 60° = 1/2 ,

sin 60° = cos 30° =

$\sqrt 3 / 2$ ,

tan 30° =

$1 / \sqrt 3$ and

tan 60° =

$\sqrt 3$ .

$U \sin g a s q u a r e d i v i d e d b y a d i a g o n a l w e f o r m t w o i s o s c e l e s r i g h t - a n g l e d t r i a n g l e s a n d c a n s e e \sin 45 ° = \cos 45 ° = 1 / \sqrt 2 a n d \tan 45 ° = 1 .$

The trigonometric functions of y = cos θ and y = cos θ.

$U \sin g t h e v a l u e s \sin 30 ° = 1 / 2 = 0.5 a n d \sin 60 ° = \sqrt 3 / 2 \approx 0.87, w e c a n d r a w u p t h e f o l l o w i n g t a b l e o f v a l u e s a n d t h e n p l o t t h e m .$


θ°	0	30	60	90	120	150	180	210	240	270	300	330	360
sin θ	0	0.5	0.87	1	0.87	0.5	0	−0.5	−0.87	−1	−0.87	−0.5	0

More points can be used to show that the shape of the graph is as shown below.

Electrical engineers and physicists call this a sine wave.

The basic formulae here all reply on the SINE graph as follows:

The sine of an angle is defined by the vertical height of a point as it rotates around a unit circle (that is, its radius is 1) measured from a horizontal line through the centre of the circle. So it cannot be bigger than 1 or less than -1.
the cosine of an angle is defined by the horizontal distance of a point as it rotates around the unit circle measured from a vertical line through the centre of the circle. It too must be in the range -1 to 1.

The surd form of trigonometric ratios:

The equilateral triangle above has sides each 2 units long and all angles at 60^o. It has been halved into 2 right – angled triangles of base 1 unit long. Using Pythagoras:

(ab)² = (ac)² + (bc)²
=> (2)² = (ac)² + (1)²
=> (ac)² = 4 – 1 = 3
and (ac) = root 3.

Using the definitions of trigonometric ratios, summarised in SOHCAHTOA:

sin 30^o = bc / ab = 1 / 2
sin 60^o = ac / ab = root 3 / 2
cos 30^o = ac / ab = root 3 / 2
cos 60^o = bc / ab = 1 /2
tan 30^o = bc / ac = 1 / root 3
tan 60 = ac / bc = root 3 / 1 = root 3

The isosceles above has equal sides of 1 unit each and the 2 complementary angles each of 45^o. Using Pythagoras’ Theorem to find side (ac):

(ac)² = (ab)² + (bc)²
(ac)² = (1)² + (1)² = 2
ac = root 2

Trigonometrical ratio:	Surd form:	Approximation
sin 30^o	1 / 2	0.5
sin 45^o	1 / root 2	0.7071
sin 60^o	root 3 / 2	0.866
cos 30^o	root 3 / 2	0.866
cos 45	1 / root 2	0.7071
cos 60^o	1 / 2	0.5
tan 30^o	1 / root 3	0.5774
tan 45^o	1
tan 60^o	root 3	1.7321

For any pair of complementary angles these 2 rules apply:

sin x = cos (90 – x)
and cos x = sin (90 – x)

sin 30^o = cos 60^o

sin 60^o = cos 30^o

sin 45^o = cos 45^o

sin 20^o = cos 70^o

cos 41^o = sin 49^o

cos 72^o = sin 18^o

sin 36^o = cos 54^o

When the angle is 30^o on a right-angled triangle, then the side opposite the 30^o angle is half the hypotenuse. This is also true for similar triangles. Sides H, A and O can be any value and O = 1 / 2 H. Therefore, sin 30^o = 0.5.

In our next class, we will be talking about Matrices and Determinant. We hope you enjoyed the class.

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Application of Surds to Trigonometric Ratios