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 .


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

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

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

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.
This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.
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.