Expert: Paul Klarreich - 5/16/2007

Question
Any help ?

a) 3sin t  - 4cos t = 2  , where 't' stands for "theta"

b) cos4x = sin 2x  

Thanks a lot !

Answer
Questioner:   Clark
Category:  Advanced Math
 
Subject:  Trigonometric Equations
Question:  Any help ?

a) 3sin t  - 4cos t = 2  , where 't' stands for "theta"

b) cos4x = sin 2x  

Thanks a lot !
 
.............................
Hi, Clark,

The second one is much easier, so I'll try that first:

b) cos(4x) = sin(2x)

cos(2(2x)) = sin 2x  

1 - 2 sin^2(2x) = sin(2x)

Let sin(2x) = y for ease of typing:

1 - 2y^2 = y

0 = 2y^2 + y - 1

That's a quadratic equation.  Factor it:

(2y - 1)(y + 1) = 0

y = 1/2  or  y = -1

sin(2x) = 1/2  or  sin(2x) = -1

For each of these, you will get your usual solutions:
.................
For the first equation:
2x = pi/6 or 5pi/6 OR 13pi/6 or 17pi/6

---- four solutions because you will now divide by 2 and get all four to be in the usual [0,2pi] interval for solutions.

x = pi/12 or 5pi/12 OR 13pi/12 or 17pi/12
.................
AND  for sin(2x) = -1:

2x = 3pi/2  or 7pi/2
x = 3pi/4  or 7pi/4
.........................................................
The second one is harder.  One way to handle it is to write:

cos t = sqrt(1 - sin^2(t))

and substitute that into the equation.  Then you have a 'radical' equation and can solve it by isolating the radical and squaring both sides.  But that solution turned out to be messy and boring.

So I think this may be more exciting and useful:  

Engineers have a way of adding up two 'sinusoidal' waves, such as  3 sin t  and  -4 cos t.  Yes, those are both sine curves, just with different amplitudes and displacement.

Any time you have

a sin t + b cos t

you can express it in terms of a single function:  A cos(t + p), where p is called the 'phase angle'.  The A (amplitude) of the sum is a Pythagorean sum of the amplitudes a,b of the original waves.

..........................
Here we go:

Let  3 sin t  - 4 cos t = A cos(t + p), where p is the phase angle.

Use a reduction formula:

A cos(t + p) = A(cos t cos p - sin t sin p) =
A cos t cos p - A sin t sin p

Now set:

A cos t cos p - A sin t sin p = 3 sin t  - 4 cos t

Equate the coefficients of  sin t and cos t:

 A cos p = - 4
- A sin p =   3

Solve for A and p.  [You have simultaneous equations.]

A = -4/cos p

- (-4/cos p) sin p =   3

4 tan p =   3
tan p = 3/4,  so  

p = arctan(3/4)  << IMPORTANT

Now if  tan p = 3/4, assuming p is in the first quadrant, (actually I am not so sure, but let it pass)

sin p = 3/5,  and  cos p = 4/5

A(4/5) = -4
A = -5   << IMPORTANT

So your left side of the equation is now:

-5 cos(t + p), where p is as above.  -- (the arctan(3/4))

And the equation says:

-5 cos(t + p) = 2

cos(t + p) = -2/5

t + p = arccos(-2/5) and
t = arccos(-2/5) - arctan(3/4)

It doesn't look great, but I can't think of anything better.