Given (A) = 5/13 and (B) = 3/5, where A and B are acute angles, find (A+B). Given (A) = 1/3 and (B) = 1/4, find (A+B).

Here are the solutions to the problems.

1. Given $\sin(A) = 5/13$ and $\cos(B) = 3/5$, where A and B are acute angles, find $\sin(A+B)$.

Step 1: Find $\cos(A)$.
Since $A$ is an acute angle, $\cos(A) > 0$.
Using the identity $\sin^2(A) + \cos^2(A) = 1$:

$$\cos^2(A) = 1 - \sin^2(A)$$ $$\cos^2(A) = 1 - \left(\frac{5}{13}\right)^2$$ $$\cos^2(A) = 1 - \frac{25}{169}$$ $$\cos^2(A) = \frac{169 - 25}{169}$$ $$\cos^2(A) = \frac{144}{169}$$ $$\cos(A) = \sqrt{\frac{144}{169}} = \frac{12}{13}$$

Step 2: Find $\sin(B)$.
Since $B$ is an acute angle, $\sin(B) > 0$.
Using the identity $\sin^2(B) + \cos^2(B) = 1$:

$$\sin^2(B) = 1 - \cos^2(B)$$ $$\sin^2(B) = 1 - \left(\frac{3}{5}\right)^2$$ $$\sin^2(B) = 1 - \frac{9}{25}$$ $$\sin^2(B) = \frac{25 - 9}{25}$$ $$\sin^2(B) = \frac{16}{25}$$ $$\sin(B) = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

Step 3: Use the sum formula for sine.
The formula for $\sin(A+B)$ is $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.

$$\sin(A+B) = \left(\frac{5}{13}\right)\left(\frac{3}{5}\right) + \left(\frac{12}{13}\right)\left(\frac{4}{5}\right)$$ $$\sin(A+B) = \frac{15}{65} + \frac{48}{65}$$ $$\sin(A+B) = \frac{15 + 48}{65}$$ $$\sin(A+B) = \frac{63}{65}$$

The value of $\sin(A+B)$ is $\boxed{\frac{63}{65}}$.

2. Given $\tan(A) = 1/3$ and $\tan(B) = 1/4$, find $\tan(A+B)$.

Step 1: Use the sum formula for tangent.
The formula for $\tan(A+B)$ is $\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$.

$$\tan(A+B) = \frac{\frac{1}{3} + \frac{1}{4}}{1 - \left(\frac{1}{3}\right)\left(\frac{1}{4}\right)}$$

Step 2: Simplify the expression.

$$\tan(A+B) = \frac{\frac{4}{12} + \frac{3}{12}}{1 - \frac{1}{12}}$$ $$\tan(A+B) = \frac{\frac{7}{12}}{\frac{12}{12} - \frac{1}{12}}$$ $$\tan(A+B) = \frac{\frac{7}{12}}{\frac{11}{12}}$$ $$\tan(A+B) = \frac{7}{12} \times \frac{12}{11}$$ $$\tan(A+B) = \frac{7}{11}$$

The value of $\tan(A+B)$ is $\boxed{\frac{7}{11}}$.

3. Solve the equation $\sin^2(x) + \cos^2(x) = 1$ for all $x$.

Step 1: Recognize the fundamental trigonometric identity.
The equation $\sin^2(x) + \cos^2(x) = 1$ is the Pythagorean identity.

Step 2: Determine the values of $x$ for which the identity holds.
This identity is true for all real numbers $x$.
The solution for $x$ is $\boxed{\text{all real numbers}}$.

4. Prove the identity $\cos(2x) = \cos^2(x) - \sin^2(x)$.

Step 1: Start with the angle sum formula for cosine.
The formula for $\cos(A+B)$ is $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$.

Step 2: Substitute $A=x$ and $B=x$ into the formula.

$$\cos(x+x) = \cos(x)\cos(x) - \sin(x)\sin(x)$$ $$\cos(2x) = \cos^2(x) - \sin^2(x)$$

The identity is proven.

5. Prove that $\tan(A-B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$.

Step 1: Express $\tan(A-B)$ in terms of sine and cosine.

$$\tan(A-B) = \frac{\sin(A-B)}{\cos(A-B)}$$

Step 2: Use the angle difference formulas for sine and cosine.

$$\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$ $$\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$$

Substitute these into the expression for $\tan(A-B)$:

$$\tan(A-B) = \frac{\sin(A)\cos(B) - \cos(A)\sin(B)}{\cos(A)\cos(B) + \sin(A)\sin(B)}$$

Step 3: Divide the numerator and denominator by $\cos(A)\cos(B)$.

$$\tan(A-B) = \frac{\frac{\sin(A)\cos(B)}{\cos(A)\cos(B)} - \frac{\cos(A)\sin(B)}{\cos(A)\cos(B)}}{\frac{\cos(A)\cos(B)}{\cos(A)\cos(B)} + \frac{\sin(A)\sin(B)}{\cos(A)\cos(B)}}$$

Step 4: Simplify the terms.
Recall that $\frac{\sin(x)}{\cos(x)} = \tan(x)$.

$$\tan(A-B) = \frac{\frac{\sin(A)}{\cos(A)} - \frac{\sin(B)}{\cos(B)}}{1 + \frac{\sin(A)}{\cos(A)}\frac{\sin(B)}{\cos(B)}}$$ $$\tan(A-B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$$

The identity is proven.

6. Given $\sin(\theta) = 4/5$ and $\theta$ is in the first quadrant, find $\sin(2\theta)$ and $\cos(2\theta)$.

Step 1: Find $\cos(\theta)$.
Since $\theta$ is in the first quadrant, $\cos(\theta) > 0$.
Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$:

$$\cos^2(\theta) = 1 - \sin^2(\theta)$$ $$\cos^2(\theta) = 1 - \left(\frac{4}{5}\right)^2$$ $$\cos^2(\theta) = 1 - \frac{16}{25}$$ $$\cos^2(\theta) = \frac{25 - 16}{25}$$ $$\cos^2(\theta) = \frac{9}{25}$$ $$\cos(\theta) = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

Step 2: Find $\sin(2\theta)$.
Use the double angle formula for sine: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.

$$\sin(2\theta) = 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)$$ $$\sin(2\theta) = 2\left(\frac{12}{25}\right)$$ $$\sin(2\theta) = \frac{24}{25}$$

Step 3: Find $\cos(2\theta)$.
Use the double angle formula for cosine: $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$.

$$\cos(2\theta) = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2$$ $$$$