Calculate the value of
`(i) sin 15^(@) " "(ii) cos 15^(@)" "(iii) tan 15^(@)`
`(iv) sin 75^(@) " "(v) cos 75^(@) " "(vi) tan 75^(@)`

Options

A. `(2-sqrt(3))/(2)`
B. `(sqrt(3)+1)/(2sqrt(2))`
C. `(sqrt(3)-1)/(-2sqrt(2))`
D. `(sqrt(3)-1)/(2sqrt(2))`

Answer

To calculate the values of \( \sin 15^\circ \), \( \cos 15^\circ \), \( \tan 15^\circ \), \( \sin 75^\circ \), \( \cos 75^\circ \), and \( \tan 75^\circ \), we can use trigonometric identities. Let's solve each part step by step.

### (i) Calculate \( \sin 15^\circ \)

We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Using the sine difference identity:

\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]

Setting \( a = 45^\circ \) and \( b = 30^\circ \):

\[ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

Substituting known values:

\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 30^\circ = \frac{1}{2} \]

Now substituting these values into the equation:

\[ \sin 15^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]

Calculating:

\[ \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### (ii) Calculate \( \cos 15^\circ \)

Using the cosine difference identity:

\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]

So,

\[ \cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \]

Substituting the known values:

\[ \cos 15^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]

Calculating:

\[ \cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### (iii) Calculate \( \tan 15^\circ \)

Using the tangent difference identity:

\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]

So,

\[ \tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

Substituting known values:

\[ \tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \]

Calculating:

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

### (iv) Calculate \( \sin 75^\circ \)

Using the sine addition identity:

\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]

So,

\[ \sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \]

Substituting known values:

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

Calculating:

\[ \sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### (v) Calculate \( \cos 75^\circ \)

Using the cosine addition identity:

\[ \cos(a + b) = \cos a \cos b - \sin a \sin b \]

So,

\[ \cos 75^\circ = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \]

Substituting known values:

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

Calculating:

\[ \cos 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### (vi) Calculate \( \tan 75^\circ \)

Using the tangent addition identity:

\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

So,

\[ \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \]

Substituting known values:

\[ \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \]

Calculating:

\[ \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Summary of Results

1. \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
2. \( \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
3. \( \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)
4. \( \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
5. \( \cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
6. \( \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)