What is an additive identities? (e.g. sin(a+b))? Then what is a double angle identities?

 

e.g. sin(2x)

 

sin(x + x) would be the right answer? But how does double angle identity applies to it?

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

 

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

http://mathworld.wolfram.com/Double-AngleFormulas.html

What a beautiful day! I just learnt about Additive & Subractive Angle Identities today. Let me recollect what I have learnt:

 

THIS IS THE FIRST SUBRACTIVE IDENTITY I HAVE LEARNT:

[math]

(cos(\alpha+\beta)-1)^2 + (sin(\alpha-\beta)-0)^2 = (cos\alpha-cos\beta)^2 + (sin\alpha-sin\beta)^2

[/math]

LEFT SIDE EXPAND:

[math]

cos^2(\alpha-\beta) - 2cos(\alpha - \beta) + 1 + sin^2(\alpha-\beta)

[/math]

THUS:

[math]

2 - 2cos(\alpha-\beta)

[/math]

RIGHT SIDE EXPAND:

[math]

cos^2{\alpha} - 2cos{\alpha}cos{\beta} + cos^2{\beta} + sin^2{\alpha} - 2sin{\alpha}sin{\beta} + sin^2{\beta}

[/math]

THUS:

[math]

2 - 2cos{\alpha}cos{\beta} - 2sin{\alpha}sin{\beta}

[/math]

COMPLETION OF BOTH SIDES:

[math]

2 - 2cos(\alpha-\beta) = 2 - 2cos{\alpha}cos{\beta} - 2sin{\alpha}sin{\beta}

[/math]

SIMPLFY:

[math]

cos(\alpha-\beta) = cos{\alpha}cos{\beta} + sin{\alpha}sin{\beta}

[/math]

 

There are 5 more additive/subractive identites but I'm not going to type all the processes right now. It took me awhile to type all the equations above.

I found out that [math]cos(2x) = 1 - 2sin^2x[/math]. How is this possible?

 

Because [imath]\cos(2x) = \cos(x+x) = \cos^2 x - \sin^2 x = (1-\sin^2 x) - \sin^2 x = 1 - 2\sin^2 x[/imath].

 

i.e. because sin² x + cos²x = 1.

One more question: How is this possible: [math]\cos(x+x) = \cos^2 x - \sin^2 x[/math]?

Well, because [imath]\cos(A+B) = \cos A \cos B - \sin A \sin B[/imath]. It's the standard compound angle formula.

[math]\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b \Rightarrow \cos \left( {a + a} \right) = \cos \left( {2a} \right) = \cos ^2 a - \sin ^2 a[/math]

Thanks TD!

Because [imath]

\cos(2x) = \cos(x+x) = \cos^2 x - \sin^2 x = (1-\sin^2 x) - \sin^2 x = 1 - 2\sin^2 x

[/imath]

 

I found out a different method to do this:

[math]

cos(a+a) = cos^2a - sin^2a = cos^2a - (1=cos^2a) = 2cos^2a-1

[/math]

 

Is the answer same as [math]1-2sin^2a[/math]?

I have provided myself with more challenging question:

 

[math]\sin(3\alpha)[/math]

 

The answer I got was [math]2\sin{\alpha}\cos^2\cos{\alpha} + \cos^2{\alpha}\sin{\alpha} - \sin^3{\alpha}[/math]

 

Is this correct?

I have provided myself with more challenging question:

 

[math]\sin(3\alpha)[/math]

 

The answer I got was [math]2\sin{\alpha}\cos^2\cos{\alpha} + \cos^2{\alpha}\sin{\alpha} - \sin^3{\alpha}[/math]

 

Is this correct?

 

Given

 

[math]\sin(\alpha+\beta)=\sin(\alpha) \cdot \cos(\beta)+\sin(\beta) \cdot \cos(\alpha)[/math]

 

and

 

[math]\cos(\alpha+\beta)=\cos(\alpha) \cdot \cos(\beta) - \sin(\alpha) \cdot \sin(\beta)[/math]

 

and

 

[math]\sin^2(\alpha)+\cos^2(\alpha)=1[/math]

 

you can figure it out...

 

[math]\sin(3\alpha)=\sin(2\alpha+\alpha)[/math]

 

[math]\sin(2\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(2\alpha)[/math]

 

[math]\sin(\alpha+\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(\alpha+\alpha)[/math]

 

[math][\sin(\alpha) \cdot \cos(\alpha) + \sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos(\alpha) \cdot \cos(\alpha) - \sin(\alpha) \cdot sin(\alpha)][/math]

 

[math]2[\sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos^2(\alpha)- \sin^2(\alpha)][/math]

 

and if...

 

[math]\sin^2(\alpha)+\cos^2(\alpha)=1[/math]

 

Then

 

[math]\cos^2(\alpha)=1-\sin^2(\alpha)[/math]

 

[math]2[\sin(\alpha) \cdot \cos^2(\alpha)]+\sin(\alpha) \cdot [1-\sin^2(\alpha)- \sin^2(\alpha)][/math]

 

[math]2[\sin(\alpha) \cdot \cos^2(\alpha)]+\sin(\alpha) \cdot [1-2\sin^2(\alpha)][/math]

 

[math]2\sin(\alpha)[\cos^2(\alpha)+1-2\sin^2(\alpha)][/math]

 

[math]2\sin(\alpha)[1-\sin^2(\alpha)+1-2\sin^2(\alpha)][/math]

 

[math]2\sin(\alpha)[2-3\sin^2(\alpha)][/math]

 

[math]4\sin(\alpha)-6\sin^3(\alpha)[/math]

 

however... from this step...

 

[math]2[\sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos^2(\alpha)- \sin^2(\alpha)][/math]

 

[math]2\sin(\alpha)\cos^2(\alpha)+ \cos^2(\alpha)\sin(\alpha) - \sin^3(\alpha)[/math]

 

can be reached (which is slightly different from your answer), but my form is simpler!

 

(last step edited by realizing my mistake...)

Is the answer same as [math]1-2sin^2a[/math']?

 

Yes, since you have 2 expressions for cos(2a), hence they must be equal.

Sorry BobbyJoeCool I mistyped my answer!!

I got the same answer as you did, but I did it in a shorter method than yours. Yours seem like forever to type!

 

THE WAY I DID WAS:

[math]

\sin(3\alpha) = \sin(2\alpha + \alpha)

[/math]

[math]

=\sin2{\alpha}\cos{\alpha} + \cos2{\alpha}\sin{\alpha}

[/math]

[math]

=(2\sin\alpha\cos\alpha)\cos\alpha + (\cos^2\alpha - \sin^2\alpha)\sin\alpha

[/math]

THEREFORE THE ANSWER IS:

[math]

\rightarrow2\sin\alpha\cos^2\alpha + \cos^2\alpha\sin\alpha - \sin^3\alpha

[/math]

You're way isn't shorter... just you don't show all the steps like I did...

 

You got (2sin•cos)cos because

 

[math]\sin(2a)=\sin(a+a)=\sin(a)\cos(a)+\sin(a)\cos(a)=2\sin(a)\cos(a)[/math]

 

and

 

[math]\cos(2a)=\cos(a+a)=\sin(a)\sin(a)-\cos(a)\cos(a)=\sin^2(a)-\cos^2(a)[/math]

 

I mearly showed those steps, so it looks longer, and I still like my final form better!

Yeah you are right BobbyJoeCool.

I HAVE A NEW QUESTION!! Give the exact value of each of the following: For example [math]\cos\frac{7\pi}{12}[/math].

 

Using the additive and subractive trigonometric identities to establish the exact value.

 

What I did was:

[math]\cos\frac{7\pi}{12} = 105[/math] degree

 

[math]\cos(\frac{\pi}{4} + \frac{\pi}{3}) = \cos\frac{\pi}{4}\cos\frac{\pi}{3} - \sin\frac{\pi}{4}\sin\frac{\pi}{3}

[/math]

[math]

(\frac{\sqrt2}{2})(\frac{1}{2}) - (\frac{\sqrt2}{2})(\frac{\sqrt3}{2})

[/math]

[math]

\frac{\sqrt2}{4} - \frac{\sqrt6}{4}

[/math]

[math]

\rightarrow\frac{\sqrt2 - \sqrt6}{4}

[/math]

 

So in the similar method above I don't know how to do this: [math]\sec\frac{-\pi}{12}[/math]??? Can anyone help me?

ugh... trig identities, hated these things so much lol

Well, [math]\sec \frac{-\pi}{12} = \frac{1}{\cos \frac{-\pi}{12}}[/math]. Now since [math]\cos \frac{-\pi}{12} = \cos \frac{\pi}{12}[/math] and you've already evaluated [math]\cos\frac{\pi}{12}[/math] you can work out the answer

I completely disagree with you dave. [math]\cos \frac{-\pi}{12}[/math] does not equal to [math] \cos \frac{\pi}{12}[/math].

 

To get [math]\sec\frac{-\pi}{12}[/math], you should have 45 degrees subract 60 degrees which would equal to minus 15 degrees. (-15 or [math]\frac{-\pi}{12}[/math]).

 

To change [math]\sec[/math], you have to get [math]\frac{1}{\cos}[/math]. Therefore,

 

[math]\frac{1}{\cos}(\frac{\pi}{4} - \frac{\pi}{3})[/math]

 

Expanding the [math]\cos[/math] subractive trig identity:

[math]

\frac{1}{\cos}(\cos\frac{\pi}{4}\cos\frac{\pi}{3} + \sin\frac{\pi}{4}\sin\frac{\pi}{3})

[/math]

 

[math]\rightarrow\frac{1}{(\frac{\sqrt2}{2})(\frac{1}{2}) + (\frac{\sqrt2}{2})(\frac{\sqrt3}{2})}

[/math]

 

[math]=\frac{1}{(\frac{\sqrt2}{4}) + (\frac{\sqrt6}{4})}[/math]

 

[math]=\frac{1}{(\frac{\sqrt2 + \sqrt6}{4})} = \frac{1}{1} \cdot \frac{4}{\sqrt2 + \sqrt6} = \frac{4}{\sqrt2 + \sqrt6}[/math]

 

Now I have to rationalize the demiantor:

[math]\frac{4}{\sqrt4 + \sqrt6} \cdot \frac{\sqrt2 - \sqrt6}{\sqrt2 - \sqrt6} = \frac{4\sqrt2 - 4\sqrt6}{2-6}

[/math]

 

[math]

\frac{4\sqrt2 - 4\sqrt6}{-4} = {-\sqrt2}{-\sqrt6}

[/math]

 

You agree with my answer or not?

[math]\cos{\tfrac{-\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}[/math]

 

[math]\cos{\tfrac{\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}[/math]

 

(Checked using TI-89 Graphing Calculator in Exact mode (AND radian mode)).

 

Think of the graph of cos... at x=0, y=1... it is symetrical as to the y-axis (as in, if you move the same distance from the y-axis in either direction, you have the same y value...)

 

Therefore, [imath]\cos{x}=\cos{-x}[/imath]

EDIT: Delete

I completely disagree with you dave. [math]\cos \frac{-\pi}{12}[/math] does not equal to [math] \cos \frac{\pi}{12}[/math].

 

To get [math]\sec\frac{-\pi}{12}[/math]' date=' you should have 45 degrees subract 60 degrees which would equal to minus 15 degrees. (-15 or [math']\frac{-\pi}{12}[/math]).

 

To change [math]\sec[/math], you have to get [math]\frac{1}{\cos}[/math]. Therefore,

 

[math]\frac{1}{\cos}(\frac{\pi}{4} - \frac{\pi}{3})[/math]

 

Expanding the [math]\cos[/math] subractive trig identity:

[math]

\frac{1}{\cos}(\cos\frac{\pi}{4}\cos\frac{\pi}{3} + \sin\frac{\pi}{4}\sin\frac{\pi}{3})

[/math]

 

[math]\rightarrow\frac{1}{(\frac{\sqrt2}{2})(\frac{1}{2}) + (\frac{\sqrt2}{2})(\frac{\sqrt3}{2})}

[/math]

 

[math]=\frac{1}{(\frac{\sqrt2}{4}) + (\frac{\sqrt6}{4})}[/math]

 

[math]=\frac{1}{(\frac{\sqrt2 + \sqrt6}{4})} = \frac{1}{1} \cdot \frac{4}{\sqrt2 + \sqrt6} = \frac{4}{\sqrt2 + \sqrt6}[/math]

 

Now I have to rationalize the demiantor:

[math]\frac{4}{\sqrt4 + \sqrt6} \cdot \frac{\sqrt2 - \sqrt6}{\sqrt2 - \sqrt6} = \frac{4\sqrt2 - 4\sqrt6}{2-6}

[/math]

 

[math]

\frac{4\sqrt2 - 4\sqrt6}{-4} = {-\sqrt2}{-\sqrt6}

[/math]

 

You agree with my answer or not?

 

Now switch pi over 4 and pi over three so you get -pi over 12... you'll get the same answer!

DUH. You're right about the cos function. I just completely forgot that it's an even function. Ok can you provide your "Now switch pi over 4 and pi over three so you get -pi over 12... you'll get the same answer!" in equations? I don't really understand where you are talking about. Thanks

Frankly you've done a lot of hard work for very little reason. You have already calculated the value of [math]\cos \frac{\pi}{12}[/math]. So now, use a series of easy arithmetic relations to get your answer:

 

[math]\sec \frac{-\pi}{12} = \frac{1}{\cos \frac{-\pi}{12}} = \frac{1}{\cos \frac{\pi}{12}}[/math]

 

Substitute in your answer and you're done.