If $$ \sin 2 A=\frac{2 \sqrt{6}}{5} $$, where $$ \hat{A}=45^{\circ} $$ Determine with the aid of a diagram and without using a calculator the value of $$ \sin A $$. If $$ \left(a^{2}+b^{2} ight) an n^{2} heta=2 a b $$ with $$ a0, b0 $$ and $$ heta \in\left(0^{\circ} ; 90^{\circ} ight) $$. Calculate the value of $$ \sin heta $$ in terms of $$ a $$ and $$ b $$. $$ 5 \quad $$ If $$ \sin 61^{\circ}=t $$, Express $$ \cos 1 $$ in terms of $$ t $$.

Question

If $$ \sin 2 A=\frac{2 \sqrt{6}}{5} $$, where $$ \hat{A}=45^{\circ} $$ Determine with the aid of a diagram and without using a calculator the value of $$ \sin A $$. If $$ \left(a^{2}+b^{2}ight) 	an n^{2} 	heta=2 a b $$ with $$ a>0, b>0 $$ and $$ 	heta \in\left(0^{\circ} ; 90^{\circ}ight) $$. Calculate the value of $$ \sin 	heta $$ in terms of $$ a $$ and $$ b $$. $$ 5 \quad $$ If $$ \sin 61^{\circ}=t $$, Express $$ \cos 1 $$ in terms of $$ t $$.

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Let's solve the problems step by step.

### Problem 1: Determine $$ \sin A $$

Given that $$ \sin 2A = \frac{2 \sqrt{6}}{5} $$ and $$ A = 45^\circ $$, we can use the double angle formula for sine:

$$
\sin 2A = 2 \sin A \cos A
$$

Substituting the known value:

$$
\frac{2 \sqrt{6}}{5} = 2 \sin A \cos A
$$

Dividing both sides by 2:

$$
\frac{\sqrt{6}}{5} = \sin A \cos A
$$

Since $$ A = 45^\circ $$, we know:

$$
\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}
$$

Thus, we can substitute:

$$
\sin A \cos A = \sin 45^\circ \cos 45^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{2}}{2}\right) = \frac{2}{4} = \frac{1}{2}
$$

Now we have:

$$
\frac{\sqrt{6}}{5} = \frac{1}{2}
$$

This is not consistent, indicating that $$ A $$ cannot be $$ 45^\circ $$ if $$ \sin 2A = \frac{2 \sqrt{6}}{5} $$. We need to find $$ A $$ such that:

$$
\sin A \cos A = \frac{\sqrt{6}}{10}
$$

To find $$ \sin A $$, we can use the identity $$ \sin^2 A + \cos^2 A = 1 $$. Let $$ \sin A = x $$ and $$ \cos A = \sqrt{1 - x^2} $$:

$$
x \sqrt{1 - x^2} = \frac{\sqrt{6}}{10}
$$

Squaring both sides:

$$
x^2 (1 - x^2) = \frac{6}{100}
$$

$$
x^2 - x^4 = \frac{3}{50}
$$

Rearranging gives us:

$$
x^4 - x^2 + \frac{3}{50} = 0
$$

Let $$ y = x^2 $$:

$$
y^2 - y + \frac{3}{50} = 0
$$

Using the quadratic formula:

$$
y = \frac{1 \pm \sqrt{1 - 4 \cdot \frac{3}{50}}}{2}
$$

Calculating the discriminant:

$$
1 - \frac{12}{50} = 1 - 0.24 = 0.76
$$

Now substituting back:

$$
y = \frac{1 \pm \sqrt{0.76}}{2}
$$

Calculating $$ \sqrt{0.76} $$:

$$
\sqrt{0.76} = \frac{2\sqrt{19}}{10}
$$

Thus:

$$
y = \frac{1 \pm \frac{2\sqrt{19}}{10}}{2}
$$

This gives us two possible values for $$ y $$, and we can find $$ \sin A $$ from $$ y = x^2 $$.

### Problem 2: Calculate $$ \sin \theta $$ in terms of $$ a $$ and $$ b $$

Given the equation:

$$
(a^2 + b^2) \tan^2 \theta = 2ab
$$

We can express $$ \tan^2 \theta $$ as:

$$
\tan^2 \theta = \frac{2ab}{a^2 + b^2}
$$

Using the identity $$ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} $$:

$$
\frac{\sin^2 \theta}{\cos^2 \theta} = \frac{2ab}{a^2 + b^2}
$$

Let $$ \sin^2 \theta = x $$ and $$ \cos^2 \theta = 1 - x $$:

$$
\frac{x}{1 - x} = \frac{2ab}{a^2 + b^2}
$$

Cross-multiplying gives:

$$
x(a^2 + b^2) = 2ab(1 - x)
$$

Rearranging:

$$
x(a^2 + b^2 + 2ab) = 2ab
$$

Thus:

$$
x = \frac{2ab}{a^2 + b^2 + 2ab}
$$

So:

$$
\sin^2 \theta = \frac{2ab}{a^2 + b^2 + 2ab}
$$

### Problem 3: Express $$ \cos 1 $$ in terms of $$ t $$

Given $$ \sin 61^\circ = t $$, we can find $$ \cos 1^\circ $$ using the identity:

$$
\cos 1^\circ = \sin(90^\circ - 1^\circ) = \sin 89^\circ
$$

Using the sine addition formula:

$$
\sin(90^\circ - x) = \cos x
$$

Thus:

$$
\cos 1^\circ = \sin 89^\circ = \sin(90^\circ - 61^\circ) = \sin 29^\circ
$$

Using the sine subtraction formula:

$$
\sin(90^\circ - x) = \cos x
$$

Thus:

$$
\cos 1^\circ = \sin(90^\circ - 61^\circ) = \sin 29^\circ
$$

Using the identity $$ \sin(90^\circ - x) = \cos x $$:

$$
\cos 1^\circ = \sqrt{1 - t^2}
$$

So, we have expressed $$ \cos 1^\circ $$ in terms of $$ t $$.

### Summary of Results
1. $$ \sin A $$ can be calculated from the quadratic equation derived.
2. $$ \sin \theta = \frac{2ab}{a^2 + b^2 + 2ab} $$.
3. $$ \cos 1^\circ = \sqrt{1 - t^2} $$.
1. $$ \sin A = \frac{2ab}{a^2 + b^2 + 2ab} $$ 2. $$ \sin \theta = \frac{2ab}{a^2 + b^2 + 2ab} $$ 3. $$ \cos 1^\circ = \sqrt{1 - t^2} $$

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