3.3 Trigonometric Identities & Equations

Consider the equation `3 cos 2x + sin x =1`.

(a) Write this equation in the form `f(x)=0`, where `f(x)=p sin^2 x+ q sin x + r`, and `p,q,r in ZZ`.

(b) Factorize `f(x)`.

(c) Write down the number of solutions of `f(x)=0`, for `0 <= x <= 2pi`.

(a) `3(1-2 sin^2 x) + sin x = 1`

`6 sin^2 x - sin x - 2 = 0 \ (p=6, q=-1, r=-2)`

(b) `(3 sin x -2)(2 sin x +1)`

(c) 4 solutions

Solve the equation `2 cos^2 x = sin 2x` for `0 <= x <= pi`, giving your answers in terms of `pi`.

Graphical solutions

for the graph of `y= 2 cos^2 x - sin 2x`

Points representing the solutions clearly indicated

1.57, 0.785

`x = pi/2 , x=pi/4`

Given that `sin x = 1/3`, where `x` is an acute angle, find the exact value of 

(a) `cos x`;

(b) `cos 2x`.

(a) `x` is an acute angle `=> cos x` is positive.

`cos^2 x + sin^2 x = 1 => cos x = sqrt (1- sin^2 x)`

`=> cos x = sqrt (1-(1/3)^2) = sqrt (8/9) = (2sqrt2)/3`

(b) `cos 2x = 1 - 2 sin^2 x = 1 - 2(1/3)^2 = 7/9`

Solve `tan(2x-5^@)=1` for `0^@ <= x <= 180^@`.

`tan^-1 1= 45^@` or equivalent

attempt to equate `2x-5^@` to their reference angle 

`2x-5^@ = 45^@, (225^@)`

`x=25^@, 115^@`

(a) Show that the equation `cos 2x = sin x` can be written in the form `2sin^2 x + sin x -1=0`.

(b) Hence, solve `cos 2x = sin x`, where `-pi <= x <= pi`.

(a) `1- 2sin^2 x = sin x`

`2 sin^2 x + sin x -1 =0`

(b) valid attempt to solve quadratic 

`(2sinx - 1)(sin x + 1)=0` OR `(-1+- sqrt(1-4(2)(-1)))/(2(2))`

recognition to solve for `sin x`

`sin x = 1/2` OR `sin x = -1`

any correct solution from `sin x = -1`

any correct solution from `sin x = 1/2`

`x = - pi/2, pi/6, (5pi)/6`

Find the least positive value of `x` for which `cos (x/2 + pi/3) = 1/sqrt 2`.

determines `pi/4` (or `45^@`) as the first quadrant (reference) angle 

attempts to solve `x/2 + pi/3 = pi/4`

`x/2 + pi/3 = pi/4 => x < 0` and so `pi/4` is rejected

`x/2 + pi/3 = 2pi - pi/4 = (7pi)/4`

`x=(17pi)/6` (must be in radians) 

It is given that `csc theta =3/2`, where `pi/2 < theta < (3pi)/2`. Find the exact value of `cot theta`.

Attempt to use `1+cot^2 theta = csc^2 theta`

`1 + cot^2 theta = 9/4`

`cot^2 theta = 5/4`

`cot theta = +- sqrt5 / 2`

`cot theta < 0` (since `csc theta > 0` puts `theta` in the second quadrant) 

`cot theta = - sqrt 5 / 2`

(a) Given that `arctan(1/5) + arctan(1/8) = arctan(1/p)`, where `p in ZZ^+`, find `p`.

(b) Hence find the value of `arctan(1/2) + arctan(1/5) + arctan(1/8)`.

(a) attempt at use of `tan(A+B) = (tan A + tan B)/(1-tanAtanB)`

`1/p = (1/5+1/8)/(1- 1/5 times 1/8)=1/3`

`p=3`

(b) `tan(arctan(1/2)+arctan(1/5)+arctan(1/8)) = (1/2+1/3)/(1-1/2 times 1/3)=1`

`arctan(1/2)+arctan(1/5)+arctan(1/8)=pi/4`

Let `f(x) = (sin 3x)/sinx - (cos3x)/cosx`.

(a) For what values of `x` does `f(x)` not exist? 

(b) Simplify the expression `(sin 3x)/sinx - (cos3x)/cosx`.

(a) `cos x = 0, sin x =0`

`x=(n pi)/2, n in ZZ`

(b) `(sin 3x cos x- cos 3x sin x)/(sin x cos x) = sin(3x-x)/(1/2 sin2x)=2`

Show that `(cos A + sin A)/(cos A - sin A)=sec 2A + tan 2A`.

`(cos A + sin A)/(cos A - sin A)=(cos A + sin A)^2/((cos A + sin A)(cos A - sin A)) = (cos^2 A + 2 cos A sin A + sin^2 A)/(cos^2 A - sin^2 A)`

`=(1+sin 2A)/(cos 2A) = sec 2A + tan 2A`