Trigonometric Problem: Solving for sin(x/2)*cos(5p/4) with Given Conditions

[diredragon]

Homework Statement

cos(x - 3p/2) = - 4/5
p <x< p/2
sin(x/2)*cos(5p/4)= ?

Homework Equations

The Attempt at a Solution

I made it as far as to determine that sinx= 4/5 and cosx = - 3/5 but can't seems to progress any further. I am looking for an easier way to find the solution without having to deal with halfangle roots and all of that. Any hints?

 

[Mentallic]

Homework Statement

cos(x - 3p/2) = - 4/5
p <x< p/2
sin(x/2)*cos(5p/4)= ?

Homework Equations

The Attempt at a Solution

I made it as far as to determine that sinx= 4/5 and cosx = - 3/5 but can't seems to progress any further. I am looking for an easier way to find the solution without having to deal with halfangle roots and all of that. Any hints?

For the first problem, what would your solution for y be to the question \cos(y)=-4/5?

For the second, are you trying to simplify \sin(x/2)*\cos(5\pi/4) into something in terms of sin(x) and cos(x)? Sometimes you are forced to deal with half angle formulas, especially when you're given a half angle.

 

[diredragon]

Actually i made an error, the question reads sin(x/3)*cos(5x/4)= ?
And for what you said y would be y= arccos(-4/5) right?

 

[Mentallic]

Actually i made an error, the question reads sin(x/3)*cos(5x/4)= ?

Ok, but that's just an expression. It's like saying, x^2+2x+1=? there's nothing to do with it from here, unless you're expected to turn it into another equal form, such as turning that example into (x+1)^2

And for what you said y would be y= arccos(-4/5) right?

Yes, right. Now let y=x-3\pi/2.

 

[diredragon]

Ok but how do i get sinx/2 ( its 2 not over 3) and cos(5x/4)? What does y = x - 3p/2 give me. I don't know y or x just siny and sinx ( through cos(A-B) being 4/5)

 

[Mentallic]

What I'm trying to get you to understand is that if
\cos(A)=B
where A and B are anything at all (avoiding impossibilities such as B>1 for example and others that I'll ignore), then it's always going to be the case that
A=\arccos(B)

so if you instead have that \cos(x-3\pi/2)=-4/5 then using the exact rules from above, where A=x-3\pi/2 and B=-4/5, what do you get? But we aren't done by that point, because we need to solve x on its own, which just requires one more simple step.

With the other problem, you're going to have to give me the exact question that you're trying to solve, word for word. At the moment, it's not a question but just an expression. I also don't know what you mean when you say "how do I get sin(x/2) and cos(5x/4)" because it makes no sense in this context. You showed me the expression \sin(x/3)*\cos(5x/4). It equals exactly that at the moment. Plug in x=0 and you'll have a value, plug in x=1 and you'll get another value. There's nothing to be done here.

 

[diredragon]

It says evaluate the x from above expression and substitute into expression sin(x/2)*(cos5x/2) to get some number. This is correct form as i made an error above. So now i see that x - 3p/2 = arccos(-4/5) but without a calculator how would i get what x is?

 

[Mentallic]

It says evaluate the x from above expression and substitute into expression sin(x/2)*(cos5x/2) to get some number. This is correct form as i made an error above. So now i see that x - 3p/2 = arccos(-4/5) but without a calculator how would i get what x is?

That was definitely an important detail that you excluded until now. The procedure for this situation would be to get the exact value of x (make x the subject), then plug it into the second expression as it is exactly, simplify a bit if possible, then use your calculator to give a numerical approximation.

The one part you need to include though is that \pi &lt; x &lt; \pi/2, but this doesn't make sense because x can't be greater than \pi and simultaneously less than \pi/2 so I'll assume you meant \pi/2 &lt; x &lt; \pi. If you're unsure. Draw a graph of y=cos(x-3pi/2) and find a few x values where its y value equals -4/5. Choose the one that lands within the desired domain of x.

 

[diredragon]

I can solve a problem using a calculator but i posted it so i can solve it by means of some simplifications and identityes, its from a recent test in which no calculators were used.

 

[Mentallic]

I can solve a problem using a calculator but i posted it so i can solve it by means of some simplifications and identityes, its from a recent test in which no calculators were used.

Then you've likely forgotten the exact question - which is also evident in the fact that you changed the question three times throughout the thread - or that you weren't expected to give a numerical answer. What you have at the moment isn't going to simplify easily at all, and this is generally the case in trigonometry.

What you may be interested in is trying to figure out what a more simple form of \cos(x-3\pi/2) is equivalent to. When sines and cosines are shifted by a factor of \pi/2, they alternate between each other.

 

[SammyS]

I can solve a problem using a calculator but i posted it so i can solve it by means of some simplifications and identityes, its from a recent test in which no calculators were used.

I suggest you give a complete statement of the complete problem, with complete sentences and using as clear mathematics as you possibly can.

 

[diredragon]

Knowing that cos(x - 3p/2) = -4/5 and (p/2)<x<p the value of the expression sin(x/2)cos(5x/2) equals what?
The solution i know must be is 82/125 but how to get there i don't know
Sorry for the mistakes before i was careless. This is complete problem read.

 

[Samy_A]

Hint:

$\sin(2x)=\sin(\frac{5}{2}x-\frac{1}{2}x)$
$\sin(3x)=\sin(\frac{5}{2}x+\frac{1}{2}x)$

 

[Ray Vickson]

Homework Statement

cos(x - 3p/2) = - 4/5
p <x< p/2
sin(x/2)*cos(5p/4)= ?

Homework Equations

The Attempt at a Solution

I made it as far as to determine that sinx= 4/5 and cosx = - 3/5 but can't seems to progress any further. I am looking for an easier way to find the solution without having to deal with halfangle roots and all of that. Any hints?

Is $p$ some other variable?

 

[SammyS]

Is $p$ some other variable?

Is $p$ some other variable?

Good question.

I'm pretty sure that Mentallic, Samy_A, and I (SammyS) have been assuming that p is indeed pi, a.k.a. π .

 

[diredragon]

No :) its pie

 

[Ray Vickson]

No :) its pie

I suppose you mean 'pi', not 'pie'. Just writing pi instead of p would make all the difference!

 

[diredragon]

I have used the property of cos(A - B) = cosAcosB + sinAsinB to get sinx = 4/5 and then by sinx^2 + cosx^2 = 1 i get cosx= -3/5.
Now for the sin(x/2)*cos(5x/2) i can substitute sin(x/2) with ((1-cosx)/2)^(1/2) and how to simplify cos((5x/2)) but that's just clearly the hearder way. Somehow this must simplify into 82/125

 

[Samy_A]

I have used the property of cos(A - B) = cosAcosB + sinAsinB to get sinx = 4/5 and then by sinx^2 + cosx^2 = 1 i get cosx= -3/5.
Now for the sin(x/2)*cos(5x/2) i can substitute sin(x/2) with ((1-cosx)/2)^(1/2) and how to simplify cos((5x/2)) but that's just clearly the hearder way. Somehow this must simplify into 82/125

Did you try this?

$\sin(2x)=\sin(\frac{5}{2}x-\frac{1}{2}x)$
$\sin(3x)=\sin(\frac{5}{2}x+\frac{1}{2}x)$

Using
$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$
$\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$
computing the two expressions in my hint and substracting them will give you $\sin(\frac{1}{2}x)\cos(\frac{5}{2}x)$ in terms of $\sin(2x)$ and $\sin(3x)$, and these two can be computed in terms of $\sin(x)=4/5$ and $\cos(x)=-3/5$.