<< Chapter < Page
  Math 1508 (lecture) readings in     Page 4 / 12
Chapter >> Page >
An isosceles right angled triangle.

If the two equal sides are of length a , then the hypotenuse, h , can be calculated as:

h 2 = a 2 + a 2 = 2 a 2 h = 2 a

So, we have:

sin ( 45 ) = opposite ( 45 ) hypotenuse = a 2 a = 1 2
cos ( 45 ) = adjacent ( 45 ) hypotenuse = a 2 a = 1 2
tan ( 45 ) = opposite ( 45 ) adjacent ( 45 ) = a a = 1

We can try something similar for 30 and 60 . We start with an equilateral triangle and we bisect one angle as shown in [link] . This gives us the right-angled triangle that we need, with one angle of 30 and one angle of 60 .

An equilateral triangle with one angle bisected.

If the equal sides are of length a , then the base is 1 2 a and the length of the vertical side, v , can be calculated as:

v 2 = a 2 - ( 1 2 a ) 2 = a 2 - 1 4 a 2 = 3 4 a 2 v = 3 2 a

So, we have:

sin ( 30 ) = opposite ( 30 ) hypotenuse = a 2 a = 1 2
cos ( 30 ) = adjacent ( 30 ) hypotenuse = 3 2 a a = 3 2
tan ( 30 ) = opposite ( 30 ) adjacent ( 30 ) = a 2 3 2 a = 1 3
sin ( 60 ) = opposite ( 60 ) hypotenuse = 3 2 a a = 3 2
cos ( 60 ) = adjacent ( 60 ) hypotenuse = a 2 a = 1 2
tan ( 60 ) = opposite ( 60 ) adjacent ( 60 ) = 3 2 a a 2 = 3

You do not have to memorise these identities if you know how to work them out.

Two useful triangles to remember

Alternate definition for tan θ

We know that tan θ is defined as: tan θ = opposite adjacent This can be written as:

tan θ = opposite adjacent × hypotenuse hypotenuse = opposite hypotenuse × hypotenuse adjacent

But, we also know that sin θ is defined as: sin θ = opposite hypotenuse and that cos θ is defined as: cos θ = adjacent hypotenuse

Therefore, we can write

tan θ = opposite hypotenuse × hypotenuse adjacent = sin θ × 1 cos θ = sin θ cos θ
tan θ can also be defined as: tan θ = sin θ cos θ

A trigonometric identity

One of the most useful results of the trigonometric functions is that they are related to each other. We have seen that tan θ can be written in terms of sin θ and cos θ . Similarly, we shall show that: sin 2 θ + cos 2 θ = 1

We shall start by considering A B C ,

We see that: sin θ = A C B C and cos θ = A B B C .

We also know from the Theorem of Pythagoras that: A B 2 + A C 2 = B C 2 .

So we can write:

sin 2 θ + cos 2 θ = A C B C 2 + A B B C 2 = A C 2 B C 2 + A B 2 B C 2 = A C 2 + A B 2 B C 2 = B C 2 B C 2 ( from Pythagoras ) = 1

Simplify using identities:

  1. tan 2 θ · cos 2 θ
  2. 1 cos 2 θ - tan 2 θ
  1. = tan 2 θ · cos 2 θ = sin 2 θ cos 2 θ · cos 2 θ = sin 2 θ
  2. = 1 cos 2 θ - tan 2 θ = 1 cos 2 θ - sin 2 θ cos 2 θ = 1 - sin 2 θ cos 2 θ = cos 2 θ cos 2 θ = 1

Prove: 1 - sin x cos x = cos x 1 + sin x

  1. LHS = 1 - sin x cos x = 1 - sin x cos x × 1 + sin x 1 + sin x = 1 - sin 2 x cos x ( 1 + sin x ) = cos 2 x cos x ( 1 + sin x ) = cos x 1 + sin x = RHS

Trigonometric identities

  1. Simplify the following using the fundamental trigonometric identities:
    1. cos θ tan θ
    2. cos 2 θ . tan 2 θ + tan 2 θ . sin 2 θ
    3. 1 - tan 2 θ . sin 2 θ
    4. 1 - sin θ . cos θ . tan θ
    5. 1 - sin 2 θ
    6. 1 - cos 2 θ cos 2 θ - cos 2 θ
  2. Prove the following:
    1. 1 + sin θ cos θ = cos θ 1 - sin θ
    2. sin 2 θ + ( cos θ - tan θ ) ( cos θ + tan θ ) = 1 - tan 2 θ
    3. ( 2 cos 2 θ - 1 ) 1 + 1 ( 1 + tan 2 θ ) = 1 - tan 2 θ 1 + tan 2 θ
    4. 1 cos θ - cos θ tan 2 θ 1 = 1
    5. 2 sin θ cos θ sin θ + cos θ = sin θ + cos θ - 1 sin θ + cos θ
    6. cos θ sin θ + tan θ · cos θ = 1 sin θ

Reduction formula

Any trigonometric function whose argument is 90 ± θ , 180 ± θ , 270 ± θ and 360 ± θ (hence - θ ) can be written simply in terms of θ . For example, you may have noticed that the cosine graph is identical to the sine graph except for a phase shift of 90 . From this we may expect that sin ( 90 + θ ) = cos θ .

Function values of 180 ± θ

Investigation : reduction formulae for function values of 180 ± θ

  1. Function Values of ( 180 - θ )
    1. In the figure P and P' lie on the circle with radius 2. OP makes an angle θ = 30 with the x -axis. P thus has coordinates ( 3 ; 1 ) . If P' is the reflection of P about the y -axis (or the line x = 0 ), use symmetry to write down the coordinates of P'.
    2. Write down values for sin θ , cos θ and tan θ .
    3. Using the coordinates for P' determine sin ( 180 - θ ) , cos ( 180 - θ ) and tan ( 180 - θ ) .
    1. From your results try and determine a relationship between the function values of ( 180 - θ ) and θ .
  2. Function values of ( 180 + θ )
    1. In the figure P and P' lie on the circle with radius 2. OP makes an angle θ = 30 with the x -axis. P thus has coordinates ( 3 ; 1 ) . P' is the inversion of P through the origin (reflection about both the x - and y -axes) and lies at an angle of 180 + θ with the x -axis. Write down the coordinates of P'.
    2. Using the coordinates for P' determine sin ( 180 + θ ) , cos ( 180 + θ ) and tan ( 180 + θ ) .
    3. From your results try and determine a relationship between the function values of ( 180 + θ ) and θ .

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Math 1508 (lecture) readings in precalculus' conversation and receive update notifications?

Ask