2. Trigonometry Exercises

By Dan Chicos. Last modified: 16 Jan 2023.

Question 1

(a) Express `5cot^2 x-2cscx+2` in terms of `cscx`.

(b) Hence, solve the equation `5cot^2 x-2cscx+2=0` for `0<=x<=2pi`.

Solution (a)

Make `cot^2 x` the subject in the identity `1+cot^2 x=csc^2 x` and substitute it in the given expression:

`1+cot^2 x=csc^2 x`

`=>` `cot^2 x=csc^2 x -1`

`5cot^2 x-2 cscx+2` `=5(csc^2 x -1)-2cscx+2`

`=5csc^2 x-5-2cscx+2`

`=5csc^2 x-2cscx-3`

Solution (b)

From (a),

`5cot^2 x-2cscx+2` `=5csc^2 x-2cscx-3,` therefore we solve `5csc^2 x-2cscx-3=0`.

Let `y=cscx` and substitute it in `5csc^2 x-2cscx-3=0`

We can solve `5y^2-2y-3=0` by factorisation or using quadratic formula.

* Using factorisation, we write `-2y=ay+by` where `a` and `b` are 2 integers such that

`a+b=-2` and `axxb=5xx(-3)=-15`. The integers must be `a=-5` and `b=3`,

therefore `-2y=ay+by`

`=-5y+3y`.

`5y^2-2y-3`

`=5y^2-5y+3y-3`

`=5y(y-1)+3(y-1)`

`=(y-1)(5y+3)`

`=0`

`y-1=0` or `5y+3=0`

`=>` `y=1` or `y=-3/5`

* Using quadratic formula `y=(-b+-sqrt(b^2-4ac))/(2a)`.

In `5y^2-2y-3=0,` `a=5,` `b=-2` and `c=-3,` doing the substitution we get

`y=(-(-2)+-sqrt((-2)^2-4xx5xx(-3)))/(2xx5)`

`=(2+-sqrt(4+60))/10`

`=(2+-8)/10`

`y=(2+8)/10=1` or `y=(2-8)/10=-6/10=-3/5`

`y=cscx`

`=>` `cscx=1` or `cscx=-3/5`

`cscx=1/sinx=1`

`=>` `sinx=1`

`=>` `x=sin^-1 1=pi/2`, `x=pi/2 in [0,2pi]`

`cscx=1/sinx=-3/5`

`=>` `sinx=-5/3`, but `-1<=sinx<=1`

`=>` `sinx=-5/3` has no real solution.

`:. x=pi/2 ~~ 1.57` is the only solution.

We can see this is indeed the case when we view the graph of `y=5cot^2 x-2cscx+2` in the region `0<=x<=2pi`.

Question 2

Find the exact value of `sin((5pi)/12)`.

Solution

It's easier to find the value of `sin((5pi)/12)` if we change the angle from radians to degrees.

`(5pi)/12 = ((5pi)/12)xx(180/pi)`

=`(5xx180)/12 = 75^@`

Since `sin75^@` is not an exact ratio, we write the angle as a sum of `45^@` and `30^@` and use the sum identity formula:

`sin75^@ = sin(45^@+30^@)`

`= sin(45^@)cos(30^@) + cos(45^@)sin(30^@)`

`= sqrt2/2xxsqrt3/2+ sqrt2/2xx1/2`

`= sqrt6/4+sqrt2/4`

`= (sqrt6 + sqrt2)/4`

`:. sin((5pi)/12) = (sqrt6 + sqrt2)/4`