If $$ \sin 9^{\circ}=\alpha $$ Find $$ \sin 18^{\circ} $$ in terms of a without using a colculator and with the aid of a diagram

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To find $$ \sin 18^{\circ} $$ in terms of $$ \alpha $$ where $$ \alpha = \sin 9^{\circ} $$, we can use the double angle formula for sine. The double angle formula states that:

$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
$$

In our case, we can set $$ \theta = 9^{\circ} $$. Therefore, we have:

$$
\sin(18^{\circ}) = \sin(2 \cdot 9^{\circ}) = 2 \sin(9^{\circ}) \cos(9^{\circ})
$$

Since $$ \sin(9^{\circ}) = \alpha $$, we can substitute this into the equation:

$$
\sin(18^{\circ}) = 2 \alpha \cos(9^{\circ})
$$

Next, we need to express $$ \cos(9^{\circ}) $$ in terms of $$ \alpha $$. We can use the Pythagorean identity:

$$
\cos^2(\theta) + \sin^2(\theta) = 1
$$

Substituting $$ \theta = 9^{\circ} $$:

$$
\cos^2(9^{\circ}) + \sin^2(9^{\circ}) = 1
$$

This gives us:

$$
\cos^2(9^{\circ}) + \alpha^2 = 1
$$

From this, we can solve for $$ \cos(9^{\circ}) $$:

$$
\cos^2(9^{\circ}) = 1 - \alpha^2
$$

Taking the square root, we have:

$$
\cos(9^{\circ}) = \sqrt{1 - \alpha^2}
$$

Now we can substitute this back into our expression for $$ \sin(18^{\circ}) $$:

$$
\sin(18^{\circ}) = 2 \alpha \sqrt{1 - \alpha^2}
$$

Thus, the final expression for $$ \sin(18^{\circ} $$ in terms of $$ \alpha $$ is:

$$
\sin(18^{\circ}) = 2 \alpha \sqrt{1 - \alpha^2}
$$
$$ \sin 18^{\circ} = 2 \alpha \sqrt{1 - \alpha^2} $$

If \( \sin 9^{\circ}=\alpha \) Find \( \sin 18^{\circ} \) in terms of a without using a colculator and with the aid of a diagram

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