Question

1.5 cos(5t)-0.8 sin(4t)=0

hey guys i got no idea how to solve this, can anyone please help me ? :)

Answer 1

\[\large{1.5\cos(5t) - 0.8\sin(4t) = 0}\]

Multiply both sides by 10, we get:

\[\large{15\cos(5t) - 8\sin(4t) = 0}\]

Any doubt ?

Answer 2

Now, lets find out this:

\[\large{\sqrt{15^2 + 8^2}}\]

Can you tell me, what is the value of the above expression ?

Answer 3

@Wazzabob you there ?

Answer 4

Okay... I would try to be as detailed as possible and would leave some things for you to solve:

I assume you have figured out \(\large{\sqrt{15^2 + 8^2}}\) which let is `a` (You must replace it with whatever value you got)

Answer 5

Now, divide both LHS and RHS by `a`:

\[\large{15\cos(5t) - 8\sin(4t) = 0}\]

\[\large{\implies {\cfrac{15}{a} \cos(5t) - \cfrac{8}{a} \sin(4t) = 0}}\]

Any doubt till this step ?

Answer 6

Just to tell you, ` a = 17`

Answer 7

Okay lets continue...

Now, this is an important and tricky step, so pay very close attention....

`If for some set of numbers 'a' and 'b'`, \(\large{a^2 + b^2 = 1}\), `then, I can replace 'a' and 'b' with : sin(u) and cos(u)`. Here:

\[\large{u = \sin^{-1}(a) = \cos^{-1}(b)}\]

Here also,

\[\large{(\cfrac{15}{17})^2 + (\cfrac{8}{17})^2 = 1}\]

So, let,

\[\large{\cfrac{15}{17} = \sin u}\]

and
\[\large{\cfrac{8}{17} = \cos u}\]

Thus, our initial equation becomes:

\[\large{\sin u *\cos(5t) - \cos u *\sin (4t) = 0}\]

Answer 8

Now recall this equation:

\[\large{\sin x \cos y = \cfrac{1}{2}(\sin(x+y) + \sin(x-y))}\]

So:

\[\large{\sin u *\cos(5t) = \cfrac{1}{2}(\sin(u+5t) + \sin(u-5t))}\]

Similarly:

\[\large{\sin(4t) * \cos(u) = \cfrac{1}{2}(\sin(4t+u) + \sin(4t-u))}\]

Answer 9

Thus, our equation becomes:

\[\large{\cfrac{1}{2}[\sin(u+5t) + \sin(u-5t)] = \cfrac{1}{2}[\sin(4t+u) + \sin(4t-u)]}\]

\[\large{\sin(u+5t) - \sin(u + 4t) = \sin(4t-u) - \sin(u-5t)}\]

Now there is an another identity:

\[\large{\sin x - \sin y = 2 \sin((x-y)/2) \cos((x+y)/2)}\]

\[\large{\sin(u+5t) - \sin(u+4t) = 2\sin(t/2) \cos(u + 9t/2)}\]

\[\large{\sin(4t-u) - \sin(u-5t) = 2\sin(9t/2 - u) \cos(-t/2)}\]

Remember that:

\[\large{\cos(-x) = \cos(x)}\]

So, our equation finally becomes:

\[\large{\sin(t/2) \cos(u + 9t/2) = \sin(9t/2 - u) \cos(t/2)}\]

Answer 10

I am terribly stuck :(

I don't think, I can help you out after this...

But, can you please recheck your question ?

Answer 11

One way to solve this is to write sin(5t) and cos(4t) in terms of sin(t) and cos(t) only and then solving the equation using a calculator...

Answer 12

Using this method, you would get:

\[\cfrac{15}{17}\cos^5 (t) - \cfrac{150}{17} \sin^{2} (t) \cos^3(t) + \\ \cfrac{75}{17} \sin^4 (t) \cos(t) = \cfrac{32}{17}\sin(t)\cos^3(t) - \cfrac{32}{17}\sin^3(t) \cos(t)\]

If I replace \(\sin(t)\) and \(\cos(t)\) by `a` and `b`, then you will have a system of equations:

\[15b^5 - 150 a^2 b^3 + 75 a^4 b = 32ab^3 - 32a^3 b\tag{1}\]
\[a^2 + b^2 = 1\tag{2}\]

You can use an advanced equation solver to solve this set of equations and get the required answer

Answer 13

As you can see... it is very very lengthy... but solvable...

That's the best I can do (at the very moment)...

Hope this helps

Answer 14

wow what a headache. This problem looks so benign and has such a pretty answer, too...

Answer 15

Yep

Answer 16

the question is good, im very bad at solving these stuff :(