trig identities

**jayshizwiz** · September 1st 2010, 12:36 PM

(of course I forgot all my high school trig...)
The problem states to find the general solution:

$sin^2 (x-30) = sin^2 2x$

I assume since both sides are sin^2, I can just say x-30 = 2x...
x=-30.

That is all I know how to do. But the answer in the book says
x1 = -30+180n
x2 = 10 + 60n

so where does the 10+60n come from and how do I know whether to add 60n or 180n or 360n...?

Thanks

**Archie Meade** · September 1st 2010, 02:16 PM

Originally Posted by jayshizwiz

(of course I forgot all my high school trig...)
The problem states to find the general solution:

$sin^2 (x-30) = sin^2 2x$

I assume since both sides are sin^2, I can just say x-30 = 2x...
x=-30.

That is all I know how to do. But the answer in the book says
x1 = -30+180n
x2 = 10 + 60n

so where does the 10+60n come from and how do I know whether to add 60n or 180n or 360n...?

Thanks

It might help to label the angles $\alpha$ and $\beta$

Then $Sin^2\alpha=Sin^2\beta$

You want the equality in terms of Sine or Cosine, so that you can say

$SinA=SinB\Rightarrow\ A=B$ or $A=180^o-B$

$CosA=CosB\Rightarrow\ A=B$ or $A=-B$

We can express $Sin^2A$ as $\frac{1}{2}\left(1-Cos(2A)\right)$

Hence

$\frac{1}{2}(1-Cos2\alpha)=\frac{1}{2}(1-Cos2\beta)$

$Cos2\alpha=Cos2\beta$

Therefore $2\alpha=2\beta$ or $2\alpha=-2\beta$

Then...

$2\alpha\ =\pm2\beta$

$2(x-30)=\pm4x$

(1)

$2x-60=4x\Rightarrow\ 2x=-60^o$

Now, do not simplify to.... $x=-30^o$ as $2x=E$

However, adding any multiple of 360 degrees to an angle is the same angle
(viewed on the circumference of a circle).
Therefore

$E=2x=-60^o+n360^o$

$x=-30^o+n180^o$

(2)

$2x-60=-4x$

$F=6x=60^o$

However, adding any multiple of 360 degrees to an angle is the same angle
(viewed on the circumference of a circle).

$6x=60^o+n360^o$

$x=10^o+n60^o$

More accurately...

$\frac{1}{2}(1-Cos2\alpha)-\frac{1}{2}(1-Cos2\beta)=0$

$\frac{1}{2}(Cos2\beta-Cos2\alpha)=0$

$Cos2\beta-Cos2\alpha=0$

Convert this to a product using the trigonometric identity $CosA-CosB=-2Sin\frac{A+B}{2}\ Sin\frac{A-B}{2}$

$\displaystyle\ Sin\left(\frac{2\alpha+2\beta}{2}\right)\ Sin\left(\frac{2\alpha-2\beta}{2}\right)=0$

Either factor can be zero since (y)(0)=0

$\displaystyle\ Sin\left(\frac{2(x-30+2x)}{2}\right)=0,\ Sin\left(\frac{2(x-30-2x)}{2}\right)=0$

$3x-30^o=n180^o,\ n=0,1,2...\Rightarrow\ 3x=30^o+n180^o\Rightarrow\ x=10^o+n60^o$

$-x-30=n180^o,\ n=0,1,2....\Rightarrow\ x=-30^o-n180^o=-30^o+n180^o$

since adding 180 degrees has the same effect as subtracting 180 degrees

**jayshizwiz** · September 1st 2010, 09:56 PM

Thank you so much for your insightful answer. I have one little question though:

Now, do not simplify to....

as

However, adding any multiple of 360 degrees to an angle is the same angle
(viewed on the circumference of a circle).
Therefore

What exactly is E? I would have gotten confused and simplified to x=-30. And since every 360 degrees to an angle is equal my answer would have been -30+360n. So could you or someone else please explain E and why I don't simplify?

Thanks again

**bigwave** · September 1st 2010, 11:08 PM

i do have a question on this
is this an identity or an equation or both
most identities i know there is no calculated value for x (altho there may be a domain)
here we can find values for x.
this was given as an identity but was solved as an equation
cranky I know...

**jayshizwiz** · September 1st 2010, 11:16 PM

ya I know it's strange but the problem says to solve for x. It's both an equation and identity i guess. Can you help me with my question??

**Archie Meade** · September 2nd 2010, 03:43 AM

Originally Posted by jayshizwiz

Thank you so much for your insightful answer. I have one little question though:

What exactly is E? I would have gotten confused and simplified to x=-30. And since every 360 degrees to an angle is equal my answer would have been -30+360n. So could you or someone else please explain E and why I don't simplify?

Thanks again

It is in fact an equation, because an identity holds for all angles.
This equation is true (RHS=LHS) only for certain angles.

The functions are periodic, hence they keep intersecting "beyond" 360 degrees.

If you calculate x from $x-30^o=2x\Rightarrow\ x=-30^o$

and $x-30^o=-2x\Rightarrow\ x=10^o$

then you are linearly obtaining a solution and you only get 2 of the solutions
(see the linear graph).

$Cos2(x-30^o)$ has twice the frequency (half the period) that $Cosx$ has.

$Cos4x$ has 4 times the frequency (a quarter of the period) that $Cosx$ has.

Hence there are multiple intersections between $0^o$ and $360^o$

That's what you need to bear in mind and that's what I was allowing for
when I used E and F.
However, it's far more accurate to use the identities quoted,
as you are then solving in terms of periodic functions.

If you are answering an exam question,
remember you will not typically have access to graphing software,
so it's important to understand this.

**jayshizwiz** · September 2nd 2010, 05:25 AM

It is in fact an equation, because an identity holds for all angles.
This equation is true (RHS=LHS) only for certain angles.

The functions are periodic, hence they keep intersecting "beyond" 360 degrees.

If you calculate x from

and

then you are linearly obtaining a solution and you only get 2 of the solutions
(see the linear graph).

has twice the frequency (half the period) that

has.

has 4 times the frequency (a quarter of the period) that

has.

Hence there are multiple intersections between

and

That's what you need to bear in mind and that's what I was allowing for
when I used E and F.
However, it's far more accurate to use the identities quoted,
as you are then solving in terms of periodic functions.

If you are answering an exam question,
remember you will not typically have access to graphing software,
so it's important to understand this.

hm...i understand what you're saying i guess, but I still don't understand why or how 2x equals the Equation. and what exactly is F and RHS LHS...

**Archie Meade** · September 2nd 2010, 05:37 AM

Originally Posted by jayshizwiz

hm...i understand what you're saying i guess, but I still don't understand why or how 2x equals the Equation. and what exactly is F and RHS LHS...

LHS means "left-hand side of the equation" and RHS means "right-hand side".

In solving for an angle of 2x, you need to write 2x=2x+n360 degrees.
If we call 2x by the label F, then F=F+n360 degrees.

Notice what happens when you divide both sides by 2.
2x=2x+n360 degrees becomes x=x+n180 degrees, which is not the same as x=x+n360.
x=x+n360 has lost the periodicity of the original function!

So, you see....it matters at what point you introduce the n360 degrees very much!

That's mainly why I "introduced" E and F, to show how the problems in finding the total solutions can arise.
It seems confusing when you try to "linearly" find the solutions.

Therefore you should use the trigonometric identities to get to the solution instead.

**jayshizwiz** · September 3rd 2010, 01:32 AM

thanks archie meade. I have a question on trig identities. I just want to go back to your original answer:

It might help to label the angles

and

Then

You want the equality in terms of Sine or Cosine, so that you can say

or

We can express

as

Hence

Therefore

or

Then...

(1)

Now, do not simplify to....

as

However, adding any multiple of 360 degrees to an angle is the same angle
(viewed on the circumference of a circle).
Therefore

(2)

However, adding any multiple of 360 degrees to an angle is the same angle
(viewed on the circumference of a circle).

More accurately...

Convert this to a product using the trigonometric identity

Either factor can be zero since (y)(0)=0

since adding 180 degrees has the same effect as subtracting 180 degrees

Are both of your answers considered trig identities, or is your first linear? I think I prefer your second answer. It's easy to understand how 0=180n.

and I have one last question at the end of your second answer:

Is it wrong to leave the answer simply as x = -30-180n? Why must you change to a +180n? (or were you just trying to show that they are the same)

**sa-ri-ga-ma** · September 3rd 2010, 03:18 AM

For n = 1,

$-30^o + 180^o = 150^o and -30^o - 180^o = -210^o$ They are identical.

But according to the conversion, angles should be measured in the counterclockwise direction. So better to write x = nπ + θ or x = nπ - θ.

You can avoid this confusion if you write the equation as

sin^(2x) - sin^2(x-30) = 0 and solve.

**Archie Meade** · September 3rd 2010, 10:43 AM

Originally Posted by jayshizwiz

thanks archie meade. I have a question on trig identities. I just want to go back to your original answer:

Are both of your answers considered trig identities, or is your first linear? I think I prefer your second answer. It's easy to understand how

$x=-30^o+n180^o$ .

Hi jayshizwiz,

Yes, the 2nd answer is one of the ways to solve these trigonometric equations.
The reason I included the method (a very awkward method) using "E" and "F",
was to give you feedback on how you were obtaining your solutions.
Hence, you'd begin to see that there is a better way.
The better way is to use the trigonometric identities.

No, the answers I gave are not trigonometric identities.
The trigonometric identities that I used are commonly listed in tables
of trigonometric identities.

The two identities I used are

$\displaystyle\ Sin^2A=\frac{1}{2}(1-Cos2A)$

$\displaystyle\ CosA-CosB=-2Sin\left(\frac{A+B}{2}\right)Sin\left(\frac{A-B}{2}\right)$

These "identities" are true or valid "in general", for all angles.

The equation that you are solving is only true for certain angles,
and your task is to discover those angles.
Look at the "periodic" attachment that I uploaded in an earlier post.

$Sin^2(x-30^o)=Sin^2(2x)$ only where those two functions intersect.

If you solve as follows....

$x-30^0=2x\Rightarrow\ -30^o=x$

$x-30^o=-2x\Rightarrow\ 3x=30^o\Rightarrow\ x=10^o$

then that's what I mean by linearly solving the equation,
but this is taking the route of the "linear" attachment I uploaded earlier.

You want to find the points of intersection of the two periodic functions,
not the points of intersection of the straight lines!

Hence, you need to use periodic methods of solution......trigonometric identities.

and I have one last question at the end of your second answer:

Is it wrong to leave the answer simply as x = -30-180n? Why must you change to a +180n? (or were you just trying to show that they are the same)

Yes, as sa-ri-ga-ma mentioned, that is convention.
In fact, you could write $-30^o=330^o$ , it's just not as convenient.

For example, starting at n=1, the angles $-30^o+n180^o$ are $150^o,\; 330^o,\; 510^o.....$

There is no need to start at n=0, since it gives the first angle as $-30^o$ , which is in fact $330^o$ .

If we left it as $-30^o-n180^0$ , the angles would be $-210^o,\; -390^o,\; -570^o...$

which are the angles $150^o,\; 330^o,\; 510^o....$

but not as easily recogniseable.
So, it's more convenient to go anticlockwise (positive angles) than clockwise (negative angles).

We could also have solved the problem using sa-ri-ga-ma's suggestion.
It looks like a lot more work, but it would be great practice!

$Sin^2(x-30^o)-Sin^2(2x)=0$

Factor....

$\left[Sin(x-30^o)+Sin(2x)\right]\left[Sin(x-30^o)-Sin(2x)\right]=0$

Either factor can be zero, but you'd need to use the trigonometric identities

$\displaystyle\ SinA+SinB=2Sin\left(\frac{A+B}{2}\right)Cos\left(\ frac{A-B}{2}\right)$

$\displaystyle\ SinA-SinB=2Cos\left(\frac{A+B}{2}\right)Sin\left(\frac{ A-B}{2}\right)$

**sa-ri-ga-ma** · September 4th 2010, 03:43 AM

There is one identity.

$sin^2{A} - sin^2{B} = sin(A+B)sin(A-B)$

**jayshizwiz** · September 5th 2010, 12:06 AM

Originally Posted by sa-ri-ga-ma

There is one identity.

$sin^2{A} - sin^2{B} = sin(A+B)sin(A-B)$

can you please show me how you got to that identity. i have looked in the internet and couldn't find that identity anywhere. here's the closest one:

cos A − cos B = −2 sin ½ (A + B) sin ½ (A − B)

**TheCoffeeMachine** · September 5th 2010, 12:14 AM

Originally Posted by jayshizwiz

can you please show me how you got to that identity. i have looked in the internet and couldn't find that identity anywhere. here's the closest one:

Look no further than our own MHF. See here.

**jayshizwiz** · September 5th 2010, 01:38 AM

I love you guys!!! Thanks again

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