EvoN1020v 12 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? 0 Share this post Link to post Share on other sites
jordan 36 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 0 Share this post Link to post Share on other sites
EvoN1020v 12 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. 0 Share this post Link to post Share on other sites
EvoN1020v 12 Posted October 24, 2005 I found out that [math]cos(2x) = 1 - 2sin^2x[/math]. How is this possible? 0 Share this post Link to post Share on other sites
Dave 246 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 sin2 x + cos2x = 1. 0 Share this post Link to post Share on other sites
EvoN1020v 12 Posted October 24, 2005 One more question: How is this possible: [math]\cos(x+x) = \cos^2 x - \sin^2 x[/math]? 0 Share this post Link to post Share on other sites
Dave 246 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. 0 Share this post Link to post Share on other sites
TD 10 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] 0 Share this post Link to post Share on other sites
EvoN1020v 12 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]? 0 Share this post Link to post Share on other sites
EvoN1020v 12 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? 0 Share this post Link to post Share on other sites
BobbyJoeCool 10 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...) 0 Share this post Link to post Share on other sites
Dave 246 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. 0 Share this post Link to post Share on other sites
EvoN1020v 12 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] 0 Share this post Link to post Share on other sites
BobbyJoeCool 10 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! 0 Share this post Link to post Share on other sites
EvoN1020v 12 Posted October 26, 2005 Yeah you are right BobbyJoeCool. 0 Share this post Link to post Share on other sites
EvoN1020v 12 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? 0 Share this post Link to post Share on other sites
CanadaAotS 12 Posted October 28, 2005 ugh... trig identities, hated these things so much lol 0 Share this post Link to post Share on other sites
Dave 246 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 0 Share this post Link to post Share on other sites
EvoN1020v 12 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? 0 Share this post Link to post Share on other sites
BobbyJoeCool 10 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] 0 Share this post Link to post Share on other sites
BobbyJoeCool 10 Posted October 28, 2005 EDIT: Delete 0 Share this post Link to post Share on other sites
BobbyJoeCool 10 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! 0 Share this post Link to post Share on other sites
EvoN1020v 12 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 0 Share this post Link to post Share on other sites
Dave 246 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. 0 Share this post Link to post Share on other sites