Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ 
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.
 A: This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.

A: 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 ).
A: 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.
A: 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
$$ \sin 18^\circ =\dfrac{\sqrt{5}-1} {4}. $$
A: 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:
$$ \sin(18^\circ)=\frac{\sqrt{5}-1}{4}\approx 0.309\ldots\tag{3}$$
