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 December 25th, 2011, 11:18 PM #1 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 trigonometric value Without using a trigonometric table, calculate the value of: $\sin$$12^{\circ}$$\cdot\sin$$48^{\circ}$$\cdot\sin $$54^{\circ}$$$
 December 26th, 2011, 01:23 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 Using product-to-sum identities, cos(60Â°) = 1/2 and cos(36Â°) - sin(18Â°) = 1/2 (which I proved here), sin(12Â°)sin(48Â°)sin(54Â°) = (1/2)(cos(36Â°) - 1/2)sin(54Â°) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ = (1/4)(2cos(36Â°)sin(54Â°) - sin(54Â°)) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ = (1/4)(1 + sin(18Â°) - sin(54Â°)) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ = (1/4)(1 - (cos(36Â°) - sin(18Â°))) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ = (1/4)(1 - 1/2) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ = 1/8. Last edited by skipjack; October 18th, 2015 at 03:16 AM.
 December 26th, 2011, 06:50 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: trigonometric value First, a bit of advice for your $\LaTeX$... For the degree symbol, use ^{\circ} and for the dot multiplication symbol use \cdot. Then use $$and$$ to enclose expressions that are taller than one character, such as variables with exponents, or fractions to automatically get the right sized bracketing symbols. This works for any bracketing symbols. I edited your post to include these. Now to the problem, and much less elegantly than [color=#00BF00]skipjack[/color]: Using a product-to-sum identity and cos(-x) = cos(x) we have: $\sin$$12^{\circ}$$\sin$$48^{\circ}$$=\frac{\cos$$\ (12-48$$^{\circ}\)-\cos$$\(12+48$$^{\circ}\)}{2}=\frac{\cos$$36^{\cir c}$$-\cos$$60^{\circ}$$}{2}=$ $\frac{1}{4}$$2\cos\(36^{\circ}$$-1\)$ Using sin(90° - x) = cos(x), we have $\sin$$54^{\circ}$$=\cos$$36^{\circ}$$$, thus: $\sin$$3\cdot18^{\circ}$$=\cos$$2\cdot18^{\circ}$$$ Using a triple-angle identity for sine on the left and a double-angle identity for cosine on the right, we have: $3\sin$$18^{\circ}$$-4\sin^3$$18^{\circ}$$=1-2\sin^2$$18^{\circ}$$$ $4\sin^3$$18^{\circ}$$-2\sin^2$$18^{\circ}$$-3\sin$$18^{\circ}$$+1=0$ Factoring yields: $$$\sin\(18^{\circ}$$-1\)$$4\sin^2\(18^{\circ}$$+2\sin$$18^{\circ}$$-1\)=0$ Knowing $0<\sin$$18^{\circ}$$<1$ we find: $\sin$$18^{\circ}$$=\frac{\sqrt{5}-1}{4}$ Using a double-angle identity for cosine, we find: $\cos$$36^{\circ}$$=\cos$$2\cdot18^{\circ}$$=1-2\sin^2$$18^{\circ}$$=1-2$$\frac{\sqrt{5}-1}{4}$$^2=$ $1-\frac{5-2\sqrt{5}+1}{8}=1-\frac{3-\sqrt{5}}{4}=\frac{1+\sqrt{5}}{4}$ Putting it all together, we find: $\sin$$12^{\circ}$$\sin$$48^{\circ}$$\sin$$54^{\cir c}$$=$ $\frac{1}{4}$$2\cos\(36^{\circ}$$-1\)\cos$$36^{\circ}$$=$ $\frac{1}{4}$$\frac{1+\sqrt{5}}{2}-1$$\frac{1+\sqrt{5}}{4}=$ $\frac{1}{32}$$\sqrt{5}-1$$$$\sqrt{5}+1$$=\frac{5-1}{32}=\frac{1}{8}$
 December 27th, 2011, 11:29 AM #4 Member   Joined: Aug 2011 From: student Posts: 55 Thanks: 0 Re: trigonometric value $\textrm{Hello! panky}$ $\textrm{Let se!}$ Using product-to-sum: $\sin(12)\sin(48)\sin(54)=\frac{1}{2}\left [ \cos(36)-\cos(60) \right ]\sin(54)$ $\sin(12)\sin(48)\sin(54)=\frac{1}{2}\cos(36)\cdot \sin(54)-\frac{1}{4}\sin(54)$ $\sin(12)\sin(48)\sin(54)=\frac{1}{4}\left [ \sin(18)+1 \right ]-\frac{1}{4}\sin(54)$ $\mathrm{\mathrm{\sin(12)\sin(48)\sin(54)=\frac{1}{ 4}+\frac{1}{4}\left [ \sin(18)-\sin(54) \right ]}$ Using triple angle identities in sin(54)=sin(3Ã—18): $\sin(12)\sin(48)\sin(54)=\frac{1}{4}+\frac{1}{4}$-2\sin(18)+4\sin^{3}(18)$$ Factorize: $\sin(12)\sin(48)\sin(54)=\frac{1}{4}+\frac{1}{4}$-2\sin(18)\cdot \cos(36)$$ $\sin(12)\sin(48)\sin(54)=\frac{1}{4}+\frac{1}{4}$-\frac{1}{2}$$ We have: $\sin(12)\sin(48)\sin(54)=\frac{1}{4}-\frac{1}{8} = \frac{1}{8}$ $\sin(12)\sin(48)\sin(54)= \frac{1}{8}$ :mrgreen: Last edited by skipjack; October 18th, 2015 at 03:07 AM.
 December 27th, 2011, 11:43 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: trigonometric value How did you know: $\sin$$18^{\circ}$$\cos$$36^{\circ}$$=\frac{1}{4}$ ?
 December 28th, 2011, 01:13 AM #6 Member   Joined: Aug 2011 From: student Posts: 55 Thanks: 0 Re: trigonometric value Yes. Let's see: We have: $\mathrm{\sin18\cdot \cos36=}$ $\mathrm{=\frac{2\sin18\cdot \cos18}{2\cos18} \cdot \cos36}$ $\mathrm{=\frac{2\sin36}{4\cos18} \cdot \cos36}$ $\mathrm{=\frac{\sin72}{4\cos18}}$ $\mathrm{=\frac{\sin(90-18 )}{4\cos18}}$ $\mathrm{=\frac{\cos(18 )}{4\cos18}}$ Thus we have : $\mathrm{=\frac14$ Last edited by skipjack; February 25th, 2016 at 04:28 AM.
 December 28th, 2011, 10:57 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 That proof is equivalent to one that appears in the post I linked to above.

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# without using tables prove that (sin12°)(sin48°)(sin54°)=1÷8

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