ROUTINE INFORMATION Hillview Secondary School
Name of School: Student Surname and Name:
Maphumulo Mthokozisi
Student Number:
207503878
Grade:
10 &11
Subject:
Mathematics
Topic:
Trigonometry , trigonometric ratios and special angles
Content /Concept Area: CAPS page number:
Cartesian plan, trigonometric ratios, special angles, quadrants and trig, function equations/ expressions. 57
Duration of Lesson:
45 minutes
Date
19 July 2018
Specific Aims: To develop understanding of special angles and meaning they attach to their scientific expression Define Trigonometric functions and plot graphs. Name the sides and angles of right angles triangles. Prepare children for being capable of dealing with trigonometric ratios and their expression in other applied maths expressions. Prepare students for future development of trigonometric function and expression involving compound angle formulae. Lesson Objectives:
Knowledge
Learners should gain
Skills Learners should be able to do
Value /Attitude Learners should acquire
knowledge and
following :
values and attitudes conducive to :
understanding of the following : Define, determine and analyse
Draw the triangle and illustrate
triangles through trigonometric
the
definitions
and their functioning.
Determine simple
Analyse
trigonometric
the
definition
function
Determine the value of the special
angles
reciprocals of
and
their
relying
on
trigonometric expression. Acknowledge ability
to
trigonometric function into
trigonometric function within
integrate the trigonometry to
exponent form
the quadrant of a circle –
special angles- their quadrant
triangle or a triangle drawn
and their relative expressions.
within a circle.
Therefore student appreciate
To analyse trig function and
skills
mark it to the suitable quadrant
deriving
being
pro-active
in
the trigonometric
function to solve problems expressed from trigonometric ratios
Describe an equation of trig.
Differentiate
Function.
Function
(Within
a
locus
point).
between equation
tri.
Develop ability Draw, label
and
and
trigonometric ratios I general
analyse
a
situation
involving both trig. Function and trig rations ‌using special angles to illustrate.
Approaches / Teaching Strategies:
Role plays. Case studies Class discussions. Questioning. Whole class-discussion Group Work Resources: OHP/ Transparency Calculators Chalkboard Newspaper Worksheet Lesson Phases: Introduction: To introduce this lesson I will show children physical objects of triangle including ruler and instruments. Basically I will be using the chalkboard instrument material. I will explain what is trigonometry briefly. Then I will request students to define three main different types of triangles. The
Cartesian plan would be used to draw the four set of quadrants and briefly illustrate the difference how the sides are named. Student require to acquire or revise knowledge of basic ratio definitions. They would also be taught to revise how to use calculator together with trigonometric equations. Students are going to be told that they need to ensure that they fully understand how to apply factorisation in trigonometric expression. Across all different books, I have noticed that there is somehow poor integration of trigonometric ratios to trigonometric equation. In this lesson plan the trigonometric equations are not going to be separated from trigonometric rations because students need to learn to deal with trigonometric expressions and their application to various functions such as trig function. This plan is to accommodate students who did not grasped the trigonometric theory from last semester. This is a plan of assisting them to face exam questions with confidence. Basically trigonometry is the study of relationship between sides and angles of a triangle, circle triangle and interaction of the radius to the circle, where the change of the size of angle and quadrant location is concerned. In grade 11 and 12 they would speak of a locus point which is fixed point from centre of a circle to the circumference of a circle. This point satisfy all conditions because it takes the form of a radius, the radius can circulate across all point of a circle. For the purpose of this lesson this implies that we are able to see 1 st, 2nd, 3rd and 4th quadrant. Then we shall use the special angles because they do not make noise in the scientific calculator when we evaluate their relationship to trigonometric expressions, trigonometric function and their ratios. The Cartesian plan appear useful when beginning to draw a right angle triangles. Later on right-angle must be separated to its circle to appear presenting special angles. Then those special angles are subject to take any form of expression imposed to them, and that truly reflect that they have been subjected under the influence of a locus point. That is why the special angles are taught together with trigonometric rations, function and equations. In two triangles below one triangle has a radius of 1 units and the other has a radius of 2 units. The special angle between each of these is 45 0. We make use of line AB and AC to make a reference of what is a locus point. A locus point of a circle take a form of a radius because it can also circulate from 1st to 4th quadrant. So students must be prepared to solve for equation, make drawings of Cartesian plane, trigonometric function and expressions, simplify and solve for equations. This requires no knowledge of Tom Dick and Hurry but it just being punctual and stick to drawings and formula sheet provided at back of the exam paper during examinations and tests.
Cartesian plane : Horizontal vs Parallel line The naming of angles and sides of 900 −∆ Right –Angled Triangle: extracted from locus of r= 1 unit
In trigonometry, sides and angles of rght angled-triangle are according to Cartesian Plane useful for competion of from the trigonometric ratios . Hence this can be used to analyse , describe and determine any equation of trigonometric expression , equation and and ratios. This can also prepare student to understand more advanced trogonometric equations such as compound angle formula. The relative smooth perfomnce of a student from more advanced trigonometric expression is linked to her background knowledge of Cartesan plane , trigonometric expressions , locus and algebraic expression. So it is better to integrate these
altogether strategically when teaching trigonometry mainly ratios , sodes and angles , function and their equations. The good thing is that they all converge into one thing called the locus point of a circle on which a ‘right- angled triangle’ is used to express their special angles. Developemnt :
Sin θ =
Opp . O = = Hyp . H
y r
Cos θ =
Adj . A = = Hyp . H
x r
Tan θ =
Opp . O = = Adj . A
y x
The above trigonometric ratios have their reciprocal ratios.
Sin θ =
Opp . O = = Hyp . H
y r
reciprocals of sin
θ
is
cosec θ=
Cos θ =
Adj . A = = Hyp . H
x r
reciprocal of
θ
is
sec =
Tan θ =
Opp . O = = Adj . A
y x
cos
reciprocal of tan
Is
cot
θ
H = O
H = A
=
r y r x
A x = O. y
Do not forget the golden rule that in any triangle ABC there are often denoted as a,b and c respectively. Some students are still totally confused what to do from the following expressions. Most of them appear to enjoy association when educators or friends illustrate bust still the side of guanine originality for these special angles expressed is duly unknown. CAPS Book Page: Challenges: Simplify these trigonometric expressions without the use of calculator 1) Sin 600.Cos 450
2) 3) 4) 5)
Sin 300 + Cos 600 Sin 450 Sin245 Cos2300
The use of special angles and sides to express these ratios in the Cartesian plane truly presents solution to students. The right angle triangle approach on which a point is drawn from the centre of a circle to radius truly make students to see what is going on with special angles expressed in trigonometric ratios. So students need to learn to produce special angles and reciprocal ratios whenever they deal with trigonometric expressions. This allows them to simplify even much complex trigonometric equations.
90
Student task: Section A:
In their groups, students must copy the drawing of a circle, where special angles of 45 0 and 300 are used to express trigonometric ratios of sin 45 and sin 30. Let students take a radius of 2 units for each trigonometric ratio. Let A be a centre of a circle and B on the right angle side and leave C to be point on the circumference for both triangles. You may draw these triangles separately. (there is no need for instrument – this require applied knowledge of geometry and trigonometry and theorem of Pythagoras). Students must discuss the theorem of Pythagoras and mark the right angle triangle. Students uses interactive chart or chalkboard guidelines to mark angles, there is no need of a protractor, this is their applied maths, however they are allowed to use calculator. Students must demonstrate the following:Observation: If trigonometric ratios of sin 45 0 and sin 300 are used, where they take a similar radius of 2 units sin ratios appear as follows.
For Sin 45: in right angled isosceles triangle:r 2 = x2 + y2 22 = x2 + x2
(x = y or x2 = y2 since triangle ABC is an isosceles triangle)
4 = 2 x2 2 x2 = 4 x2 = 2 x=
√2
If evaluated correctly sin 45 =
√2 2
, this is equivalent to
1 . √2
(you may rationalise
denominator to verify). Therefore Sin 45 =
√2 2
=
1 √2
when r =
√2
remain constant even if r= 2 units. Since sin 45 =
1 . We say these values ae equivalent. √2
For second triangle ABC, where A is a centre, B is a point of right angle and C on the circumference of the 1st quadrant.
a scalene ¿ angled triangle ∆ ABC, A is centre of a circle, ABC is right angle of a scalene right
In
angled triangle ABC and angle ACB is angle on the circumference, where radius meets the point BC
scalene ∆ ABC given angle CAB = 300.
placed to right angle. In
∴∈∆ ABC:-
¿ BCA + ¿ CAB + ¿ ABC
= 1800 - sum of a triangle
¿ BCA + 300+ 900 = 1800 = 1800 – (1200) = 600.
¿ BCA This means cos 600 =
x r
But given, r = 2
x 2
So
r = 2(
= cos 600, r = 2 cos 600
1 ¿ 2
= 1 units.
You should use sin 30 to determine the value of y but also you should use the cos of 30 to determine the value of x directly. But we have to manipulate the formula with an interest of demonstrating students what going on behind the invention of trigonometric ratios. Theorem of Pythagoras: x2 + y2 = r2 x2+ (1)2 = 22 x2 = 22 -(1)2 = 4 -1
∴ X2 = 3 X=
√3
Therefore where r = 2 units, the sin of 30 degrees is
1 2
1 . But when r = 1 units sin 30 = Sin 30 0 = 2
Therefore Where r= 2, sin 30 =
1 2
, also where r = 1 sin 30 =
1 2
.
Planned Questions: Return to Classwork: Challenges Simplify the following expressions by using special angles and not a calculator. 1) 2) 3) 4) 5) i) ii) iii)
Sin 600.Cos 450 Sin 300 + Cos 600 Sin 450 Sin245 Cos2300 What is Cartesian plane and how did it applied to your practical? How are sides and angles named after the Pythagoras Theorem? a) Give at least three examples of special angle and also b) give any of their corresponding trigonometric ratio for each. c) Draw and locate the quadrant of the circle on a Cartesian plane: comment on the value of sin (120). (Hint: Sin 120 = Sin (180 - β ¿ Students are expected to respond as follows:a) 300 , 450, 600 b) Here answers may range from student to student depending what trigonometric ratio he may have chosen (to pick up). c) Student must demonstrate knowledge of CAST interpretation. The answer is not interest at a moment. But answer is Sin (120) = Sin (180 – 60) = sin 60 Sin 60 =
√3 2
.
You may also verify these values on the calculator.
The isosceles triangles and right angled triangle of trigonometric ratios and functions are useful. Students are encouraged to make copy of the example below to use in their exam.
Teacher’s task: Consolidation and Conclusion To consolidate a lesson. For r = 2 and r = 1 values of sin 45, and cos 45 remain constant. Therefore students can enjoy integrating understanding of locus to trigonometric ratios, special angles and sides and trigonometric function and equation. Section B Teachers task: Given a triangle ABC, in which radius is 1 units …. Determine the trigonometric rations of sin 45 0 and sin 300. In this triangle keep triangles same but let them remain across different Cartesian plane. So you may produce a set of two drawings.
Assessment: Classwork and Homework Using the Cartesian plane draw a triangle in which A is at centre of a circle. Now mark the angle of 300against the x-axis in an anticlockwise direction. Joint the point of an angle to produce a radius and of AC. Produce a vertical line from x-axis to join line AC at point so that ( X: Y ) is crated > Therefore ABC is a right angle triangle on which angle ABC is a right angle , CAB = 30 0an and BCA is 60 degrees. NB the sum of a triangle = 180 0. Now ignoring the value of angles of a triangle, use x, y and r to express the following: Tan
θ
=
sinθ cos θ
hint look at the locus and apply the theorem of Pythagoras. (Here we are
testing skills of students to be able to draw a triangle rough sketch as soon as they can, to meet exam requirements). Split the LHS and RHS to prove that they are equal. (It simplify). Now use the angles to prove the very same theory, but this time around you will apply the special angles knowledge as follows:
Tan
30 =
sin 30 cos 30
us r = 2 for scalene angle triangle (please do not use r = 1 unit) or rather
they are given.
Conclusion: Trigonometry looks at relationship between the sides and angles. We often use the Cartesian plane together with right- angles triangle to illustrate relationship that exist between sides. We employ the locus plane drawn in a circle to demonstrate how trigonometric function and angles are subject to change across different quadrants (CAST diagram). The special angles, 30 0, 450 , 600, 900, 1200, 1500, 1800, etc. are used to demonstrate to determine the actual values of trigonometric ratios, their reciprocals and special angles. There are three main set of trigonometric ratios and each ratio has its reciprocal. We have seen cos, sin and tan known as the trigonometric ratio as well as the sec, cosec and cot known as the reciprocal trigonometric functions respectively. NB In the word trigonometry: tri- means three (3) which reflect that we are dealing with triangles to express relationship between triangles and sides expressed in a circle on which Cartesian plane is used to express relationship across four different quadrants. The word CAST diagram is an abbreviation which was adopted by mathematicians to demonstrate change of the value of trig ratio as radius or locus point moves from 0 0 to say 1500, 300 or 2400 and 450 etc. This lesson plan concluded that it is possible to deal with trig ratios, their reciprocals, special angles and sides and trigonometric expressions and function all in one plane. This asserts that students must be prepared to get any kind of equation or trig function and at
any time. In as much as these revisions overlap we assume that it is due to application of locus delimitation in a circle where trig –ratios and their reciprocals are inserted within the right angled triangle to take different expressions across different location of circle diagram or locus points (a circle). Grade 10, 11 and 12 specifically need to realise that Pythagoras theorem and Cartesian plane is becoming more important for invention of trigonometric expressions and their function. Sometimes it involves algebraic expressions, exponential forms and even logarithmic expressions. Geometry is essential for one to be able to deal with trigonometry and one can argue that 1 in 5 sum expressed in trigonometry required the knowledge of geometry mainly analytic and equations on which Pythagoras theorem applies directly or indirectly. When will students being able to see if triangle is scalene right angled triangle, isosceles or equilateral if he has poor knowledge of geometry. When will students see if r is hypotenuse and s is distance whereas y is vertical opposite side if he has no understanding of geometry and Cartesian plane? . When will students understand the rotational angle that keep the radius alive from 1st to 4th quadrant of a CAST diagram. All these points truly reflect that when we are dealing with trigonometric ratios, special angles and sides, we will need understanding of geometry. The end