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Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .

$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\ &=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\ &=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\ &=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\ \end{align}$

$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\ &=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\ &=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\ &=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\ \end{align}$

$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\ &=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\ \end{align}$

is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$

I have studied maths upto $12$th grade.

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    $\begingroup$ You can use their Taylor expansions $\endgroup$ – Rol Apr 20 '15 at 13:38
  • $\begingroup$ I haven't studied advanced calculus like taylor series, $\endgroup$ – R K Apr 20 '15 at 13:40
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    $\begingroup$ @RK Any answers you get are likely to be better if you specify which tools you do have available. (For the second, anyway, one could use that $18^{\circ}$ is one-quarter of $72^{\circ}$, the well-known value of a trig function at the latter, and a half-angle identity.) $\endgroup$ – Travis Apr 20 '15 at 13:53
  • $\begingroup$ i don't understand what u mean by tools, i can't use calculators i have to find it by pen and paper. I know only trignometric values of $0,30,45,60 \quad \text{and} \quad 90$ $\endgroup$ – R K Apr 20 '15 at 13:56
  • $\begingroup$ tools as in mathematical theory $\endgroup$ – MCT Apr 20 '15 at 14:02
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This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.

enter image description here

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  • $\begingroup$ i don't know what to do, do i have to find the length of side of the triangle with respect to side of a pentagon $\endgroup$ – R K Apr 20 '15 at 14:25
  • $\begingroup$ Exactly. The easy part is finding half of the side; the hard part is finding the diagonal relative to a side. Try Googling "pentagon golden ratio" for ideas. see a related post math.stackexchange.com/questions/763128/… $\endgroup$ – John Joy Apr 20 '15 at 14:32
  • $\begingroup$ have a look at mathuprising.comlu.com/images/pentagon.png <-- gives answer away $\endgroup$ – John Joy Apr 20 '15 at 19:51
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Please refer to : derivation of sin 18 on this page

Once you know sin 18, you can find sin 9, cos 9 etc. by using half angle formulas.
In brief:

sin 72° = 2 sin 36° cos 36° by the double angle relationship. sin 72° = 4 sin 18° cos 18° (1 - 2sin^2 18°) by the double angle relationship, again. cos 18° = 4 sin 18° cos 18° (1 - 2sin^2 18°)
sin 72° = cos 18°. 1 = 4 sin 18° (1 - 2sin^2 18°)

Let x = sin 18°, this is known as 1 = 4*x(1-2x^2) substitution 8*x^3-4*x+1 = 0 A product is zero only when one of its factors is zero. 8x^3-4x+1 = (2*x-1)(4*x^2+2*x-1)=0 (2*x-1)=0 implies x= ½=sin 30° > sin 18° ; Since we know sin is increasing on [0°,90°]. x = (-2 ± \sqrt{(4 + 4•4•1))/8} So we must solve the other factor, = (-2 ± \sqrt{20})/8 using the quadratic formula. = (-2 ± \sqrt{4}\sqrt{5})/8 = (-1 ± \sqrt{5})/4 But the sin 18° > 0, so it cannot be negative. sin 18° = (\sqrt{5}-1)/4 Hence the middle root is the one we want.

Here at the bottom of the page referred above you will see a comment about how to find sin 1 also. From sin 1 you can find sin (1/2) and note that 23 = (22 +(1/2)) + (1/2). But 22 + ( 1/2 ) is 1/2 of ( 45 ).

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Hint: in this table you have the values of: $$ \sin 18°\qquad \sin 3° $$ From these you can find: $$ \sin 21°=\sin(18°+3°) \qquad \sin 69°=\sin(90°-21°) $$ All these are constructible numbers, i.e. real numbers that we can express using square roots (and the other arithmetic operations).

For $\tan 23°$ you can note that $69°=3 \times 23°$and use the formula: $$ \tan 3 \alpha=\dfrac{3\tan \alpha-\tan^3 \alpha}{1-3\tan^2 \alpha} $$ But this gives a cubic equation and this means that the number $ \tan (23°)$ is an algebraic number but it is not constructible. If you know how to solve a cubic you can find a finite expression for $ \tan (23°)$ , but if you does not know, you can only find an approximate value as shown in other answers.

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I do not know if you know multiple angle trig formulas.

Let $ A = 18 ^\circ. $ In a right angled triangle if acute angles are $ 2A= 36 ^\circ, \,3A= 54^\circ$,

$ \sin 2 A = \cos 3A $

$ 2 \sin A \cos A = 4 \cos^3 A -3 \cos A $

simplifying and solving for $ \sin A $ gives you

$$ \sin 18^\circ =\dfrac{\sqrt{5}-1} {4}. $$

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You may exploit: $$ \sin(60^\circ)=\frac{1}{2}\sqrt{3},\quad \sin(18^\circ)=\frac{1}{4}\left(\sqrt{5}-1\right),\quad \tan(22^\circ 30')=\sqrt{2}-1$$ then use some form of interpolation. For instance, in a neighbourhood of $x=\frac{\pi}{3}$: $$ \sin(x)=\frac{1}{2}\sqrt{3}+\frac{1}{2}\left(x-\frac{\pi}{3}\right)-\frac{\sqrt{3}}{4}\left(x-\frac{\pi}{3}\right)^2-\frac{1}{12}\left(x-\frac{\pi}{3}\right)^3+\ldots $$ so by taking $x=\frac{\pi}{3}+\frac{9\pi}{180}$ we have: $$ \sin(69^\circ) \approx \frac{\sqrt{3}}{2}+\frac{\pi}{40}-\frac{\pi^2\sqrt{3}}{1600}=0.933\ldots\tag{1}$$ and in a neighbourhood of $x=\frac{\pi}{8}$ we have: $$ \tan(x)=(\sqrt{2}-1)+(4-2\sqrt{2})\left(x-\frac{\pi}{8}\right)+(6\sqrt{2}-8)\left(x-\frac{\pi}{8}\right)^2+\ldots $$ so by taking $x=\frac{\pi}{8}+\frac{\pi}{360}$ we have: $$ \tan(23^\circ) \approx \sqrt{2}-1+(4-2\sqrt{2})\frac{\pi}{360}=0.424\ldots \tag{2}$$ At last, we have: $$ \sin(18^\circ)=\frac{\sqrt{5}-1}{4}\approx 0.309\ldots\tag{3}$$

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For Sin 69∘, Exact value may be determined as follows:

+45..............45.....................+2

+22.5............67.5...................+2

+11.25............78.75................+2

-5.625............73.125................-2

-2.8125............70.3125.............+2

-1.40625............68.90625.........+2

0.703125............69.609375........-2

-0.3515625..........69.2578125......-2

-0.17578125........69.08203125............+2

-0.087890625........68.9941406250........+2

0.043945313........69.0380859375........-2

-0.021972656........69.0161132813........-2

-0.010986328........69.0051269531........+2

-0.005493164........68.99963378906250.....+2

Hint for table: First column is half of earlier (called as Central) with sign towards 69∘; second is cumulative of the first column; and the third is division current Central with earlier Central. If we follow above pattern infinitely, the sum will be 69. However, we can see the repeating pattern in third column (+2, +2, -2, -2); hence this is the exact value.

Now write the third column top-to-down approach as:$$+2+2+2-2\overline{+2+2-2-2} $$.

In the next step just replace 2 by √(2; the result will be : $$√(2+√(2+√(2-√(2\overline{+√(2+√(2-√(2-√(2} $$.

Sin 69∘ will be half of this (closing brackets are collapsed by "]"). Therefore, Sin 69∘ = $$√(2+√(2+√(2-√(2\overline{+√(2+√(2-√(2-√(2}]/2 $$

Above method is called as Precise-Rewritten method. You can find exact trigonometric value of all integer angles using this method.

I apologize above bad formatting and discourage to explain all angles here.

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  • $\begingroup$ Request to visitors and down voters: May you assist me to copy edit that document? I have not idea to use all of these. I cannot format as online forum requires formatting the mathematics. ** If someone assisted** me for copy edit on Precise-Rewritten method (and other new methods), every scholar may know the new idea for new method for exact trigonometric values. My (Breaking Classical Rules in Trigonometry- Mission 2050) un-skill on mathematics formating discouraging me to expose all of those new idea. $\endgroup$ – Bhava Nath Dahal Nov 7 '16 at 4:57

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