Computer `sin 75^(@) , cos 75^(@) and tan 15 ^(@),` from the functions of `30^(@) and 45 ^(@)`

Answer

To compute \( \sin 75^\circ \), \( \cos 75^\circ \), and \( \tan 15^\circ \) using the values of \( 30^\circ \) and \( 45^\circ \), we can use the sine and cosine addition formulas. Here’s the step-by-step solution:

### Step 1: Calculate \( \sin 75^\circ \)

We can express \( 75^\circ \) as \( 45^\circ + 30^\circ \).

Using the sine addition formula:
\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]
Let \( x = 45^\circ \) and \( y = 30^\circ \):
\[ \sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \]

Now, substituting the known values:
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 30^\circ = \frac{1}{2} \)

So,
\[ \sin 75^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]
\[ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### Step 2: Calculate \( \cos 75^\circ \)

Using the cosine addition formula:
\[ \cos(x + y) = \cos x \cos y - \sin x \sin y \]
Let \( x = 45^\circ \) and \( y = 30^\circ \):
\[ \cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \]

Substituting the known values:
\[ \cos 75^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]
\[ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### Step 3: Calculate \( \tan 15^\circ \)

We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \).

Using the tangent subtraction formula:
\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]
Let \( x = 45^\circ \) and \( y = 30^\circ \):
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

Substituting the known values:
- \( \tan 45^\circ = 1 \)
- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)

So,
\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]
\[ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \]

Multiplying the numerator and denominator by \( \sqrt{3} \):
\[ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Final Answers:
- \( \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
- \( \cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
- \( \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)