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θ 0 30 45 60 90 180
tan θ 0 1 3 1 3 0

Now that we have graphs for sin θ and cos θ , there is an easy way to visualise the tangent graph. Let us look back at our definitions of sin θ and cos θ for a right-angled triangle.

sin θ cos θ = opposite hypotenuse adjacent hypotenuse = opposite adjacent = tan θ

This is the first of an important set of equations called trigonometric identities . An identity is an equation, which holds true for any value which is put into it. In this case we have shown that

tan θ = sin θ cos θ

for any value of θ .

So we know that for values of θ for which sin θ = 0 , we must also have tan θ = 0 . Also, if cos θ = 0 our value of tan θ is undefined as we cannot divide by 0. The graph is shown in [link] . The dashed vertical lines are at the values of θ where tan θ is not defined.

The graph of tan θ .

Functions of the form y = a tan ( x ) + q

In the figure below is an example of a function of the form y = a tan ( x ) + q .

The graph of 2 tan θ + 1 .

Functions of the form y = a tan ( θ ) + q :

  1. On the same set of axes, plot the following graphs:
    1. a ( θ ) = tan θ - 2
    2. b ( θ ) = tan θ - 1
    3. c ( θ ) = tan θ
    4. d ( θ ) = tan θ + 1
    5. e ( θ ) = tan θ + 2
    Use your results to deduce the effect of q .
  2. On the same set of axes, plot the following graphs:
    1. f ( θ ) = - 2 · tan θ
    2. g ( θ ) = - 1 · tan θ
    3. h ( θ ) = 0 · tan θ
    4. j ( θ ) = 1 · tan θ
    5. k ( θ ) = 2 · tan θ
    Use your results to deduce the effect of a .

You should have found that the value of a affects the steepness of each of the branches. The larger the absolute magnitude of a , the quicker the branches approach their asymptotes, the values where they are not defined. Negative a values switch the direction of the branches. You should have also found that the value of q affects the vertical shift as for sin θ and cos θ . These different properties are summarised in [link] .

Table summarising general shapes and positions of graphs of functions of the form y = a tan ( x ) + q .
a > 0 a < 0
q > 0
q < 0

Domain and range

The domain of f ( θ ) = a tan ( θ ) + q is all the values of θ such that cos θ is not equal to 0. We have already seen that when cos θ = 0 , tan θ = sin θ cos θ is undefined, as we have division by zero. We know that cos θ = 0 for all θ = 90 + 180 n , where n is an integer. So the domain of f ( θ ) = a tan ( θ ) + q is all values of θ , except the values θ = 90 + 180 n .

The range of f ( θ ) = a tan θ + q is { f ( θ ) : f ( θ ) ( - , ) } .


The y -intercept, y i n t , of f ( θ ) = a tan ( x ) + q is again simply the value of f ( θ ) at θ = 0 .

y i n t = f ( 0 ) = a tan ( 0 ) + q = a ( 0 ) + q = q


As θ approaches 90 , tan θ approaches infinity. But as θ is undefined at 90 , θ can only approach 90 , but never equal it. Thus the tan θ curve gets closer and closer to the line θ = 90 , without ever touching it. Thus the line θ = 90 is an asymptote of tan θ . tan θ also has asymptotes at θ = 90 + 180 n , where n is an integer.

Graphs of trigonometric functions

  1. Using your knowldge of the effects of a and q , sketch each of the following graphs, without using a table of values, for θ [ 0 ; 360 ]
    1. y = 2 sin θ
    2. y = - 4 cos θ
    3. y = - 2 cos θ + 1
    4. y = sin θ - 3
    5. y = tan θ - 2
    6. y = 2 cos θ - 1
  2. Give the equations of each of the following graphs:

The following presentation summarises what you have learnt in this chapter.


  • We can define three trigonometric functions for right angled triangles: sine (sin), cosine (cos) and tangent (tan).
  • Each of these functions have a reciprocal: cosecant (cosec), secant (sec) and cotangent (cot).
  • We can use the principles of solving equations and the trigonometric functions to help us solve simple trigonometric equations.
  • We can solve problems in two dimensions that involve right angled triangles.
  • For some special angles, we can easily find the values of sin, cos and tan.
  • We can extend the definitions of the trigonometric functions to any angle.
  • Trigonometry is used to help us solve problems in 2-dimensions, such as finding the height of a building.
  • We can draw graphs for sin, cos and tan

End of chapter exercises

  1. Calculate the unknown lengths
  2. In the triangle P Q R , P R = 20  cm, Q R = 22  cm and P R ^ Q = 30 . The perpendicular line from P to Q R intersects Q R at X . Calculate
    1. the length X R ,
    2. the length P X , and
    3. the angle Q P ^ X
  3. A ladder of length 15 m is resting against a wall, the base of the ladder is 5 m from the wall. Find the angle between the wall and the ladder?
  4. A ladder of length 25 m is resting against a wall, the ladder makes an angle 37 to the wall. Find the distance between the wall and the base of the ladder?
  5. In the following triangle find the angle A B ^ C
  6. In the following triangle find the length of side C D
  7. A ( 5 ; 0 ) and B ( 11 ; 4 ) . Find the angle between the line through A and B and the x-axis.
  8. C ( 0 ; - 13 ) and D ( - 12 ; 14 ) . Find the angle between the line through C and D and the y-axis.
  9. A 5 m ladder is placed 2 m from the wall. What is the angle the ladder makes with the wall?
  10. Given the points: E(5;0), F(6;2) and G(8;-2), find angle F E ^ G .
  11. An isosceles triangle has sides 9 cm , 9 cm and 2 cm . Find the size of the smallest angle of the triangle.
  12. A right-angled triangle has hypotenuse 13 mm . Find the length of the other two sides if one of the angles of the triangle is 50 .
  13. One of the angles of a rhombus ( rhombus - A four-sided polygon, each of whose sides is of equal length) with perimeter 20 cm is 30 .
    1. Find the sides of the rhombus.
    2. Find the length of both diagonals.
  14. Captain Hook was sailing towards a lighthouse with a height of 10 m .
    1. If the top of the lighthouse is 30 m away, what is the angle of elevation of the boat to the nearest integer?
    2. If the boat moves another 7 m towards the lighthouse, what is the new angle of elevation of the boat to the nearest integer?
  15. (Tricky) A triangle with angles 40 , 40 and 100 has a perimeter of 20 cm . Find the length of each side of the triangle.

Questions & Answers

how do you translate this in Algebraic Expressions
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Maths grade 10 rought draft. OpenStax CNX. Sep 29, 2011 Download for free at http://cnx.org/content/col11363/1.1
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