• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  • Level: GCSE
  • Subject: Maths
  • Word count: 2542

Graphs of Sin x, Cos x; and Tan x

Extracts from this document...

Introduction

image00.png

Graphs of sinx°, cosx° and tanx°

Here are the sketch graphs of the trigonometric functions f(x) = sinx°, f(x) = cosx° and f(x) = tanx°.

image03.pngimage04.pngimage30.png

You may be asked to draw or sketch these graphs in your exam. Try to remember what they look like, and follow these tips:

If you are asked to draw or plot the graph, you will need to use your calculator to generate the y-values. For example, if you were asked to plot the graph of f(x) = sinx° for 0° x 360° , you would use you calculator to find sin0°, sin10°, sin 20°, ...., sin 360° and then plot these values on the graph paper.

Plotting a trigonometric graph is time-consuming and it is therefore more likely that you will be asked to sketch the graph. However, even if you think that you remember what the graph looks like, your calculator can be used to check. For example, sin0° = 0 and cos0° = 1, so you have the starting points of the graphs. Tan90° has no value (your calculator will display an error message), so you know that the graph cannot cross the line x = 90°.

image26.png

Transformations of graphs; y = asinbx° and y = acosbx°

Remember

Given a graph, f(x):

  • The transformation af(x) causes a stretch, parallel to the y-axis with a scale factor of a.
  • The transformation f(bx) causes a stretch, parallel to the x-axis with a scale factor of image62.png.
...read more.

Middle

.

Remember

This is the circle property which is the most difficult to spot. Look out for a triangle with one of its vertices resting on the point of contact of the tangent.

image40.png

The angle between a tangent and a chord is equal to the angle made by that chord in the alternate segment.

image41.png

Question 1

What is the size of:

  1. angle x?
  2. angle y?

image42.png

The Answer

  1. Did you get x = 60°? Well done!
  2. Did you get y = 80°? Well done! You remembered that the angles in a triangle add up to 180°.
  • Make sure that you learn these circle properties. If you are asked to find the angles in a circle, you will then be able to see which of them apply to the question.
  • Do not be afraid to find the sizes of other angles first. It is not always possible to find the required angle immediately!

image01.png

Congruency

If two shapes are congruent, then they are identical in shape and size.

Question 1

Which of the following shapes are congruent?

image43.png

The Answer

Did you get the following pairs?


A and G
D and I
E and J
C and H

Well done! Remember that shapes can be congruent even if one of them has been rotated (as in A and G) or reflected (as in C and H).

The symbol means 'is congruent to'.

Two triangles are congruent if one of the following conditions applies:

The three sides of the first triangle are equal to the three sides of the second triangle.(SSS)

image44.png

Two sides of the first triangle are equal to two sides of the second triangle and the included angle is equal.(SAS)

image45.png

Two angles in the first triangle are equal to two angles in the second triangle and one (similarly located) side is equal. (AAS)

image46.png

In a right angled triangle, the hypotenuse and one other side in the first triangle are equal to the hypotenuse and the corresponding side in the second triangle. (RHS)

image47.png

Question 2

For each of the following pairs of triangles, state whether they are congruent. If they are, give a reason for your answer (SSS, SAS, AAS or RHS).

Pair 1

image48.png

Pair 2

image49.png

Pair 3

image50.png

The Answer

  1. Yes. RHS
  2. Yes. SSS
  3. Did you say no? Well done! You spotted that the side of length 7cm was not in the same position on both triangles. Therefore it is not AAS.

image26.png

...read more.

Conclusion

image79.png

Question 1

Write down the three ways of describing the vector if the arrow is pointing in the opposite direction.

image80.png

The Answer

Did you get image81.png, -a and image82.png? Well done!

Teacher's Note

If not, remember that the arrow describes the direction, so in this case, the vector is from B to A. If we move 'backwards' along a vector, it becomes negative, so a becomes -a.

image84.png

image26.png

Vector 'arithmetic'

Equal vectors

If two vectors have the same magnitude and direction, then they are equal.

image85.png

Adding vectors

image86.png

Vector image87.pngfollowed by vector image88.pngrepresents a movement from P to R. image89.png

image90.png

image91.png

Subtracting vectors

image92.png

Vector image11.png, followed by a backwards movement along image93.png, is equivalent to a movement from X to Z.image94.png

image95.png

image96.png

Multiplication by a scalar

image97.png

e.g. image98.png

image99.png

Question 1

If x = image100.png, y = image101.pngand z = image06.png, find:

  1. -y
  2. x - y
  3. 2x + 3z

The Answer

  1. image07.png. Did you remember to change the signs?
  2. image08.png
  3. image09.png

image10.png

To travel from X to Z, it is possible to move along vector image11.png, followed by image12.png. It is also possible to go directly along image13.png.

image13.pngis therefore known as the resultant of image11.pngand image12.png

Geometric problems

Question 1

image15.png

Write as single vectors:

  1. f + g
  2. a + b
  3. e - b - a

The Answer

  1. e
  2. -c Did you remember the minus sign?
  3. -d

Remember

Two vectors are equal if they have the same magnitude and direction, regardless of where they are on the page.

Question 2

Triangles ABC and XYZ are equilateral. X is the midpoint of AB, Y is the midpoint of BC and Z is the midpoint of AC. image16.png= a , image13.png= b and image17.png= c. Express each of the following in terms of a, b and c.

  1. image11.png
  2. image12.png
  3. image18.png
  4. image19.png
  5. image20.png

image22.png

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Fencing problem.

    occur as if one misprint or incorrect digit is missing a total error in the final output can obliterate the value that shall be obtained. It shall be more manageable if it was simplified. I have shown the simplest form that I can represent it in below (the difference that

  2. Geography Investigation: Residential Areas

    In this hypothesis I will use every street that I have surveyed. For the first part of the hypothesis I will only be using the 'external' questionnaire - this is the survey I did myself of the area. Table 6 Index Of Decay Street Sarum Hill Bounty Road Penrith Road

  1. HL type 1 portfolio on the koch snowflake

    The difference in the area between two successive triangles is represented by the dark triangles in the figure below. Stage 1 Stage 2 This difference is demonstrated in step 1 as. For Stage 4, It is also demonstrated in step 3 that the as 'n' increases it corresponds to a

  2. Beyond Pythagoras

    84 96 54 4 180 150 5 330 B Input 2n� 1 2 2 16 3 54 4 128 5 250 Rule = 2n� A-B Input A-B Output 1ST Difference 2nd Difference 1 6-2 4 10 6 2 30-16 14 16 6 3 84-54 30 22 6 4 180-128 52

  1. Beyond Pythagoras

    5 6 12 2 5 12 13 30 30 3 7 24 25 84 56 4 9 40 41 180 90 5 11 60 61 330 132 6 13 84 85 546 182 ( Based on the above table, I can work out the expressions for: 1.

  2. Beyond Pythagoras

    which can be expanded to 6n2 + 6n Longest side (c) This side is always 3 more than b in this family, so the formula is 6n2 + 6n +3 Area (A) The formula for the area (A) of any of the triangles is 1/2 ab, which can be written

  1. Investigate the affects of the surface area: volume ratio on the cooling of an ...

    in a small beaker, each unit of volume inside the beaker has on average less surface area in contact with the outside. So the heat from the water inside the small beaker will be lost faster because more of the water is closer to the surface and the distance that

  2. Beyond Pythagoras.

    Term 5: 'h' = 2n2 + 2n + 1 = 2 � 5� + 2 � 5 + 1 = 2 � 25 + 10 + 1 = 50 + 11 = 61 Term 4: 'h' = 2n2 + 2n + 1 = 2 � 4� + 2 � 4

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work