Difficult Problem.... 12(sin 2x) + 5(cos 2x) = 13[sin (2x + a)] where a = Tan^-1 (12/5).?

Question

Yes true or false.

Answer 1

Just leave it like that.

Answer 2

12sin(2x) + 5cos(2x) = 13[sin(2x + a)]

First, let's expand sin(2x + a) using the sine addition identity. A reminder that
sin(c + d) = sin(c)cos(d) + sin(d)cos(c), so we have

12sin(2x) + 5cos(2x) = 13[sin(2x)cos(a) + cos(2x)sin(a)]

Now, all we have to do is solve for two values:
1) cos(a), which is also equal to cos[tan^(-1)(12/5)]
2) sin(a), which is also equal to sin[tan^(-1)(12/5)]

Let's solve for cos(a) first.
Let t = tan^(-1)(12/5). Taking the tan of both sides,

tan(t) = 12/5.

By SOHCAHTOA, tan(t) = opp/adj = 12/5, so
opp = 12
adj = 5
So by Pythagoras, hyp = sqrt(5^2 + 12^2) = sqrt(169) = 13.

We want cos(t) = cos[tan^(-1)(12/5)] = adj/hyp = 5/13, so
cos(a) = 5/13.

It follows that sin(a) = opp/hyp = 12/13.

So now, our new equation is

12sin(2x) + 5cos(2x) = 13[ sin(2x)[5/13] + cos(2x)[12/13] ]

Distributing the 13 outside of the brackets will get rid of both fractions, leaving us with

12sin(2x) + 5cos(2x) = 5sin(2x) + 12cos(2x)

Now, we can group like terms upon moving everything to the left hand side.

7sin(2x) - 7cos(2x) = 0

Move the 7cos(2x) to the right hand side,

7sin(2x) = 7cos(2x)

And now, dividing both sides by 7cos(2x), we get

sin(2x)/cos(2x) = 1, or
tan(2x) = 1

Assuming a restricted interval of 0 <= x < 2pi,
tan is equal to 1 at the points pi/4 and 5pi/4. Note that we have to double our values because of the 2 in front of the x; that is, we have to add 2pi to each of pi/4 and 5pi/4. This gives us:

2x = {pi/4, 5pi/4, pi/4 + 2pi, 5pi/4 + 2pi}. Simplifying this,

2x = {pi/4, 5pi/4, 9pi/4, 13pi/4}. Multiplying both sides by (1/2), we get

x = {pi/8, 5pi/8, 9pi/8, 13pi/8}

This is false, because if this were true, we would get all real numbers as a solution. As you can see, we do not.

Answer 3

is this true or false?