<s>[INST] 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.
 [/INST] 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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