# Trigonometric Identities

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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?

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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:

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

LEFT SIDE EXPAND:

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

THUS:

$2 - 2cos(\alpha-\beta)$

RIGHT SIDE EXPAND:

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

THUS:

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

COMPLETION OF BOTH SIDES:

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

SIMPLFY:

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

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.

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

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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.

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

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

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$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b \Rightarrow \cos \left( {a + a} \right) = \cos \left( {2a} \right) = \cos ^2 a - \sin ^2 a$

Thanks TD!

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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:

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

Is the answer same as $1-2sin^2a$?

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I have provided myself with more challenging question:

$\sin(3\alpha)$

The answer I got was $2\sin{\alpha}\cos^2\cos{\alpha} + \cos^2{\alpha}\sin{\alpha} - \sin^3{\alpha}$

Is this correct?

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I have provided myself with more challenging question:

$\sin(3\alpha)$

The answer I got was $2\sin{\alpha}\cos^2\cos{\alpha} + \cos^2{\alpha}\sin{\alpha} - \sin^3{\alpha}$

Is this correct?

Given

$\sin(\alpha+\beta)=\sin(\alpha) \cdot \cos(\beta)+\sin(\beta) \cdot \cos(\alpha)$

and

$\cos(\alpha+\beta)=\cos(\alpha) \cdot \cos(\beta) - \sin(\alpha) \cdot \sin(\beta)$

and

$\sin^2(\alpha)+\cos^2(\alpha)=1$

you can figure it out...

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

$\sin(2\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(2\alpha)$

$\sin(\alpha+\alpha) \cdot \cos(\alpha)+\sin(\alpha) \cdot \cos(\alpha+\alpha)$

$[\sin(\alpha) \cdot \cos(\alpha) + \sin(\alpha) \cdot \cos(\alpha)] \cdot \cos(\alpha)+\sin(\alpha) \cdot [\cos(\alpha) \cdot \cos(\alpha) - \sin(\alpha) \cdot sin(\alpha)]$

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

and if...

$\sin^2(\alpha)+\cos^2(\alpha)=1$

Then

$\cos^2(\alpha)=1-\sin^2(\alpha)$

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

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

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

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

$2\sin(\alpha)[2-3\sin^2(\alpha)]$

$4\sin(\alpha)-6\sin^3(\alpha)$

however... from this step...

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

$2\sin(\alpha)\cos^2(\alpha)+ \cos^2(\alpha)\sin(\alpha) - \sin^3(\alpha)$

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 $1-2sin^2a[/math']? Yes, since you have 2 expressions for cos(2a), hence they must be equal. ##### Link to post ##### Share on other sites Sorry BobbyJoeCool I mistyped my answer!! I got the same answer as you did, but I did it in a shorter method than yours. Yours seem like forever to type! THE WAY I DID WAS: [math] \sin(3\alpha) = \sin(2\alpha + \alpha)$

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

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

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

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

You got (2sin•cos)cos because

$\sin(2a)=\sin(a+a)=\sin(a)\cos(a)+\sin(a)\cos(a)=2\sin(a)\cos(a)$

and

$\cos(2a)=\cos(a+a)=\sin(a)\sin(a)-\cos(a)\cos(a)=\sin^2(a)-\cos^2(a)$

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

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Yeah you are right BobbyJoeCool.

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

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

What I did was:

$\cos\frac{7\pi}{12} = 105$ degree

$\cos(\frac{\pi}{4} + \frac{\pi}{3}) = \cos\frac{\pi}{4}\cos\frac{\pi}{3} - \sin\frac{\pi}{4}\sin\frac{\pi}{3}$

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

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

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

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

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

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

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I completely disagree with you dave. $\cos \frac{-\pi}{12}$ does not equal to $\cos \frac{\pi}{12}$.

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

To change $\sec$, you have to get $\frac{1}{\cos}$. Therefore,

$\frac{1}{\cos}(\frac{\pi}{4} - \frac{\pi}{3})$

Expanding the $\cos$ subractive trig identity:

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

$\rightarrow\frac{1}{(\frac{\sqrt2}{2})(\frac{1}{2}) + (\frac{\sqrt2}{2})(\frac{\sqrt3}{2})}$

$=\frac{1}{(\frac{\sqrt2}{4}) + (\frac{\sqrt6}{4})}$

$=\frac{1}{(\frac{\sqrt2 + \sqrt6}{4})} = \frac{1}{1} \cdot \frac{4}{\sqrt2 + \sqrt6} = \frac{4}{\sqrt2 + \sqrt6}$

Now I have to rationalize the demiantor:

$\frac{4}{\sqrt4 + \sqrt6} \cdot \frac{\sqrt2 - \sqrt6}{\sqrt2 - \sqrt6} = \frac{4\sqrt2 - 4\sqrt6}{2-6}$

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

You agree with my answer or not?

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$\cos{\tfrac{-\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$

$\cos{\tfrac{\pi}{12}}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$

(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

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I completely disagree with you dave. $\cos \frac{-\pi}{12}$ does not equal to $\cos \frac{\pi}{12}$.

To get $\sec\frac{-\pi}{12}$' 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 $\sec$, you have to get $\frac{1}{\cos}$. Therefore,

$\frac{1}{\cos}(\frac{\pi}{4} - \frac{\pi}{3})$

Expanding the $\cos$ subractive trig identity:

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

$\rightarrow\frac{1}{(\frac{\sqrt2}{2})(\frac{1}{2}) + (\frac{\sqrt2}{2})(\frac{\sqrt3}{2})}$

$=\frac{1}{(\frac{\sqrt2}{4}) + (\frac{\sqrt6}{4})}$

$=\frac{1}{(\frac{\sqrt2 + \sqrt6}{4})} = \frac{1}{1} \cdot \frac{4}{\sqrt2 + \sqrt6} = \frac{4}{\sqrt2 + \sqrt6}$

Now I have to rationalize the demiantor:

$\frac{4}{\sqrt4 + \sqrt6} \cdot \frac{\sqrt2 - \sqrt6}{\sqrt2 - \sqrt6} = \frac{4\sqrt2 - 4\sqrt6}{2-6}$

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

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!

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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

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

$\sec \frac{-\pi}{12} = \frac{1}{\cos \frac{-\pi}{12}} = \frac{1}{\cos \frac{\pi}{12}}$

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