EvoN1020v 12 Posted October 24, 2005 Share Posted October 24, 2005 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? Link to post Share on other sites

jordan 36 Posted October 24, 2005 Share Posted October 24, 2005 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 Link to post Share on other sites

EvoN1020v 12 Posted October 24, 2005 Author Share Posted October 24, 2005 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. Link to post Share on other sites

EvoN1020v 12 Posted October 24, 2005 Author Share Posted October 24, 2005 I found out that [math]cos(2x) = 1 - 2sin^2x[/math]. How is this possible? Link to post Share on other sites

Dave 251 Posted October 24, 2005 Share Posted October 24, 2005 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 sin^{2} x + cos^{2}x = 1. Link to post Share on other sites

EvoN1020v 12 Posted October 24, 2005 Author Share Posted October 24, 2005 One more question: How is this possible: [math]\cos(x+x) = \cos^2 x - \sin^2 x[/math]? Link to post Share on other sites

Dave 251 Posted October 24, 2005 Share Posted October 24, 2005 Well, because [imath]\cos(A+B) = \cos A \cos B - \sin A \sin B[/imath]. It's the standard compound angle formula. Link to post Share on other sites

TD 10 Posted October 24, 2005 Share Posted October 24, 2005 [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] Link to post Share on other sites

EvoN1020v 12 Posted October 24, 2005 Author Share Posted October 24, 2005 Thanks TD! Link to post Share on other sites

EvoN1020v 12 Posted October 24, 2005 Author Share Posted October 24, 2005 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]? Link to post Share on other sites

EvoN1020v 12 Posted October 25, 2005 Author Share Posted October 25, 2005 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? Link to post Share on other sites

BobbyJoeCool 10 Posted October 25, 2005 Share Posted October 25, 2005 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...) Link to post Share on other sites

Dave 251 Posted October 25, 2005 Share Posted October 25, 2005 Is the answer same as [math]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

EvoN1020v 12 Posted October 25, 2005 Author Share Posted October 25, 2005 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] Link to post Share on other sites

BobbyJoeCool 10 Posted October 26, 2005 Share Posted October 26, 2005 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! Link to post Share on other sites

EvoN1020v 12 Posted October 26, 2005 Author Share Posted October 26, 2005 Yeah you are right BobbyJoeCool. Link to post Share on other sites

EvoN1020v 12 Posted October 28, 2005 Author Share Posted October 28, 2005 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? Link to post Share on other sites

CanadaAotS 12 Posted October 28, 2005 Share Posted October 28, 2005 ugh... trig identities, hated these things so much lol Link to post Share on other sites

Dave 251 Posted October 28, 2005 Share Posted October 28, 2005 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 Link to post Share on other sites

EvoN1020v 12 Posted October 28, 2005 Author Share Posted October 28, 2005 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? Link to post Share on other sites

BobbyJoeCool 10 Posted October 28, 2005 Share Posted October 28, 2005 [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] Link to post Share on other sites

BobbyJoeCool 10 Posted October 28, 2005 Share Posted October 28, 2005 EDIT: Delete Link to post Share on other sites

BobbyJoeCool 10 Posted October 28, 2005 Share Posted October 28, 2005 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! Link to post Share on other sites

EvoN1020v 12 Posted October 28, 2005 Author Share Posted October 28, 2005 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 Link to post Share on other sites

Dave 251 Posted October 28, 2005 Share Posted October 28, 2005 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. Link to post Share on other sites

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