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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 .

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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