James Gray
University of Edinburgh
Applicable Mathematics (Foundation) af0
6: Further Trigonometry
Recap
soh cah toa
`tan theta = (sin theta)/(cos theta)`
`sin^2 theta = (sin theta)^2` and `cos^2 theta = (cos theta)^2`
So applying Pythagoras's Theorem to the diagram above we get the new formula
`sin^2theta + cos^2 theta =1`
Note
From symmetries in graphs
`sin (-theta) = -sin(theta)`
`cos (-theta) = cos (theta)`
For example
`sin (-45^circ) = -1/sqrt(2) = -sin 45^circ`
`cos (-30^circ) = sqrt(3)/2 = cos(30^circ)`
6.1 Addition Formulae
Notice that you cannot just add values of trig functions to get the corresponding angles. For example
`sin 30^circ + sin 30^circ = 1/2 + 1/2 = 1`
However
`sin 60^circ = sqrt(3) / 2`.
So
`sin 60^circ != sin 30^circ + sin 30^circ`.
Therefore we need much more complicated formulae for addition. These are
`cos(alpha +beta)=cos alpha cos beta - sin alpha sin beta`
Subtraction
Likewise we need complicated formulae for subtraction. These don't need to be remembered because they can be derived from the above formulae.
`sin(alpha-beta) = sin(alpha + (-beta))`
`=sin alpha cos (-beta) + cos alpha sin (-beta)`
`=sin alpha cos beta - cos alpha sin beta`.
`cos(alpha - beta)=cos (alpha +(-beta))`
`=cos alpha cos(-beta) - sin alpha sin (-beta)`
`=cos alpha cos beta + sin alpha sin beta`.
Example
Q: Find the exact value of `sin 15^circ`
A: `60^circ -45^circ = 15^circ`
So `sin 15^circ = sin(60^circ -45^circ)`
`= sin 60^circ cos 45^circ - cos 60^circ sin 45^circ`
`= sqrt(3)/2 1/2 - 1/2 1/sqrt(2) = (sqrt(3) - 1)/(2sqrt(2))`
6.2 Identities
At the start of this chapter we saw that
`sin^2 theta + cos^2 theta -=1`.
Notice we have used the symbol "`-=`" rather than just "`=`". The symbol "`-=`" is used to say that this formula is true for all values of `theta`. Such a statement is called an identity. Compare that with the formula
`sin theta = 1/2`
which is only true for certain values of `theta`.
Another example of an identity is
`(x+1)(x-1) -= x^2-1`
because this is true for all values of `x`. However
`x^2-1=0`
is only true when `x=1` or `x=-1`.
Proving Trig Identities
To show that an equation is an identity we must start with one side and then derive the other.
Example
Q: Prove that `(cos(alpha - beta))/(cos alpha cos beta) -= 1 + tan alpha tan beta`.
A: `LHS = (cos(alpha - beta))/(cos alpha cos beta) `
`=(cos alpha cos beta + sin alpha sin beta)/(cos alpha cos beta)`
`= 1 + (sin alpha sin beta)/(cos alpha cos beta)`
`=1 + tan alpha tan beta`
`=RHS`
6.3 Double Angle Formulae
`sin 2alpha -= 2 sin alpha cos alpha`
`cos 2 alpha -= cos^2 alpha - sin^2 alpha`
Why? For `sin 2 alpha`
`LHS = sin 2 alpha = sin (alpha + alpha)`
`=sin alpha cos alpha + cos alpha sin alpha`
`=2sin alpha cos alpha = RHS`
The working is similar to show the other formula.
There are other ways of writing the formula for `cos 2 alpha`
`cos 2 alpha -= 2 cos^2 alpha -1`
`cos 2 alpha -= 1 - 2 sin^2 alpha`
Why? `LHS = cos 2 alpha = cos^2 alpha - sin^2 alpha`
But `sin^2 alpha + cos^2 alpha = 1`,
so `sin^2 alpha = 1 - cos^2 alpha`.
Therefore
`cos^2 alpha - sin^2 alpha = cos^2 alpha - (1 - cos^2 alpha)`
`=cos^2 - 1 + cos ^2 alpha`
`=2 cos^2 alpha -1 = RHS`
Again the working is similar for the other formula.
Example
Q: Show that `sin4theta = 4 sin theta cos^3 theta - 4 sin^3 theta cos theta`.
A: `LHS = sin 4 theta`
`=sin (2 theta + 2theta)`
`=sin 2theta cos 2theta + cos 2theta sin 2 theta`
`=2 sin2theta cos2theta`
`=2(2sin theta cos theta)(cos^2 theta - sin^2 theta)`
`=4 sin theta cos^3 theta - 4 sin^3 theta cos theta`
6.4 Solving Trig Equations
Example
Q: Solve `2cos^2 x - 3 cos x + 1 = 0` for `0 le x le 360^circ`.
A: Write `c` for `cos x` and factorise:
`2c^2 - 3c + 1 =0`
`(2c-1)(c-1) =0`
So `c=1/2` or `c=1`.
Therefore `cos x = 1/2` or `cos x =1`.
Now use method from section 3.4.
For `cos x = 1/2` the associated acute angle is `cos^(-1) 1/2 = 60^circ` and hence `x=60^circ, 300^circ`.
For `cos x =1` the associated acute angle is `cos^(-1) 1 = 0` and hence `x = 0^circ, 360^circ`.
Therefore the solutions of `2cos^2 x - 3 cos x + 1 = 0` are `x = 0^circ, 60^circ, 300^circ, 360^circ`.
Using the Double Angle Formulae:
Example
Q: Solve `sin 2x - cos x =0` for `0^circ le x le 360^circ`.
A: Use `sin 2x = 2 sin x cos x` to get
`2 sin x cos x - cos x = 0`
`cos x (2sin x -1) =0`
So `cos x = 0` or `sin x =1/2`. Using the method of associated acute angles in each case we find that `x=30^circ, 90^circ, 150^circ, 270^circ`.