(d) If { \sin 12^ \circ} = M$}$, Write The Following In Terms Of { M$}$ 1. { \tan 12^{\circ $}$2. { \sin 114^{\circ}$}$3. { \sin 6^{\circ} \cos 6^{\circ}$} 4. \[ 4. \[ 4. \[ \cos^2 6^{\circ} - \sin^2

Expressing Trigonometric Functions in Terms of $\sin 12^{\circ}$

Introduction

In this article, we will explore the relationship between various trigonometric functions and express them in terms of $\sin 12^{\circ}$. We will use the given value of $\sin 12^{\circ} = m$ to derive the expressions for $\tan 12^{\circ}$, $\sin 114^{\circ}$, $\sin 6^{\circ} \cos 6^{\circ}$, and $\cos^2 6^{\circ} - \sin^2 6^{\circ}$.

1. Expressing $\tan 12^{\circ}$ in Terms of $\sin 12^{\circ}$

To express $\tan 12^{\circ}$ in terms of $\sin 12^{\circ}$, we can use the definition of tangent as the ratio of sine and cosine.

$\tan 12^{\circ} = \frac{\sin 12^{\circ}}{\cos 12^{\circ}}$

We can rewrite $\cos 12^{\circ}$ in terms of $\sin 12^{\circ}$ using the Pythagorean identity:

$\cos^2 12^{\circ} + \sin^2 12^{\circ} = 1$

$\cos 12^{\circ} = \sqrt{1 - \sin^2 12^{\circ}}$

Substituting this expression into the definition of tangent, we get:

$\tan 12^{\circ} = \frac{\sin 12^{\circ}}{\sqrt{1 - \sin^2 12^{\circ}}}$

2. Expressing $\sin 114^{\circ}$ in Terms of $\sin 12^{\circ}$

To express $\sin 114^{\circ}$ in terms of $\sin 12^{\circ}$, we can use the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 114^{\circ} = \sin (90^{\circ} + 24^{\circ})$

Using the angle addition formula, we get:

$\sin 114^{\circ} = \sin 90^{\circ} \cos 24^{\circ} + \cos 90^{\circ} \sin 24^{\circ}$

Since $\sin 90^{\circ} = 1$ and $\cos 90^{\circ} = 0$, we can simplify the expression to:

$\sin 114^{\circ} = \cos 24^{\circ}$

We can rewrite $\cos 24^{\circ}$ in terms of $\sin 12^{\circ}$ using the double-angle formula for cosine:

$\cos 2a = 2\cos^2 a - 1$

In this case, we have:

$\cos 24^{\circ} = \cos 2(12^{\circ})$

Using the double-angle formula, we get:

$\cos 24^{\circ} = 2\cos^2 12^{\circ} - 1$

We can rewrite $\cos^2 12^{\circ}$ in terms of $\sin 12^{\circ}$ using the Pythagorean identity:

$\cos^2 12^{\circ} = 1 - \sin^2 12^{\circ}$

Substituting this expression into the double-angle formula, we get:

$\cos 24^{\circ} = 2(1 - \sin^2 12^{\circ}) - 1$

Simplifying the expression, we get:

$\cos 24^{\circ} = 1 - 2\sin^2 12^{\circ}$

Therefore, we can express $\sin 114^{\circ}$ in terms of $\sin 12^{\circ}$ as:

$\sin 114^{\circ} = 1 - 2\sin^2 12^{\circ}$

3. Expressing $\sin 6^{\circ} \cos 6^{\circ}$ in Terms of $\sin 12^{\circ}$

To express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$, we can use the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 12^{\circ} = \sin (6^{\circ} + 6^{\circ})$

Using the angle addition formula, we get:

$\sin 12^{\circ} = \sin 6^{\circ} \cos 6^{\circ} + \cos 6^{\circ} \sin 6^{\circ}$

Simplifying the expression, we get:

$\sin 12^{\circ} = 2\sin 6^{\circ} \cos 6^{\circ}$

We can rewrite $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ by dividing both sides of the equation by 2:

$\sin 6^{\circ} \cos 6^{\circ} = \frac{\sin 12^{\circ}}{2}$

4. Expressing $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in Terms of $\sin 12^{\circ}$

To express $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$, we can use the Pythagorean identity:

$\cos^2 a + \sin^2 a = 1$

In this case, we have:

$\cos^2 6^{\circ} + \sin^2 6^{\circ} = 1$

Subtracting $\sin^2 6^{\circ}$ from both sides of the equation, we get:

$\cos^2 6^{\circ} = 1 - \sin^2 6^{\circ}$

We can rewrite $\sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ using the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 12^{\circ} = \sin (6^{\circ} + 6^{\circ})$

Using the angle addition formula, we get:

$\sin 12^{\circ} = \sin 6^{\circ} \cos 6^{\circ} + \cos 6^{\circ} \sin 6^{\circ}$

Simplifying the expression, we get:

$\sin 12^{\circ} = 2\sin 6^{\circ} \cos 6^{\circ}$

We can rewrite $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ by dividing both sides of the equation by 2:

$\sin 6^{\circ} \cos 6^{\circ} = \frac{\sin 12^{\circ}}{2}$

Substituting this expression into the Pythagorean identity, we get:

$\cos^2 6^{\circ} = 1 - \left(\frac{\sin 12^{\circ}}{2}\right)^2$

Simplifying the expression, we get:

$\cos^2 6^{\circ} = 1 - \frac{\sin^2 12^{\circ}}{4}$

We can rewrite $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ by subtracting $\sin^2 6^{\circ}$ from both sides of the equation:

$\cos^2 6^{\circ} - \sin^2 6^{\circ} = 1 - \frac{\sin^2 12^{\circ}}{4} - \sin^2 6^{\circ}$

We can rewrite $\sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ using the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 12^{\circ} = \sin (6^{\circ} + 6^{\circ})$

Using the angle addition formula, we get:

$\sin 12^{\circ} = \sin 6^{\circ} \cos 6^{\circ} + \cos 6^{\circ} \sin 6^{\circ}$

Simplifying the expression, we get:

$\sin 12^{\circ} = 2\sin 6^{\circ} \cos 6^{\circ}$

We can rewrite $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ by dividing both sides of the equation by 2:

$\sin 6^{\circ} \cos 6^{\circ} = \frac{\sin 12^{\circ}}{2}$

Substituting this expression into the Pythagorean identity, we get:

$\sin^2 6^\circ} = \left(\frac{\sin 12
**Q&A Expressing Trigonometric Functions in Terms of $\sin 12^{\circ$**

Introduction

In our previous article, we explored the relationship between various trigonometric functions and expressed them in terms of $\sin 12^{\circ}$. We derived expressions for $\tan 12^{\circ}$, $\sin 114^{\circ}$, $\sin 6^{\circ} \cos 6^{\circ}$, and $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$. In this article, we will answer some frequently asked questions related to these expressions.

Q: What is the relationship between $\tan 12^{\circ}$ and $\sin 12^{\circ}$?

A: The relationship between $\tan 12^{\circ}$ and $\sin 12^{\circ}$ is given by the expression:

$\tan 12^{\circ} = \frac{\sin 12^{\circ}}{\sqrt{1 - \sin^2 12^{\circ}}}$

This expression shows that $\tan 12^{\circ}$ is equal to $\sin 12^{\circ}$ divided by the square root of $1 - \sin^2 12^{\circ}$.

Q: How can we express $\sin 114^{\circ}$ in terms of $\sin 12^{\circ}$?

A: We can express $\sin 114^{\circ}$ in terms of $\sin 12^{\circ}$ using the angle addition formula for sine:

$\sin 114^{\circ} = \sin (90^{\circ} + 24^{\circ})$

Using the angle addition formula, we get:

$\sin 114^{\circ} = \sin 90^{\circ} \cos 24^{\circ} + \cos 90^{\circ} \sin 24^{\circ}$

Since $\sin 90^{\circ} = 1$ and $\cos 90^{\circ} = 0$, we can simplify the expression to:

$\sin 114^{\circ} = \cos 24^{\circ}$

We can rewrite $\cos 24^{\circ}$ in terms of $\sin 12^{\circ}$ using the double-angle formula for cosine:

$\cos 2a = 2\cos^2 a - 1$

In this case, we have:

$\cos 24^{\circ} = \cos 2(12^{\circ})$

Using the double-angle formula, we get:

$\cos 24^{\circ} = 2\cos^2 12^{\circ} - 1$

We can rewrite $\cos^2 12^{\circ}$ in terms of $\sin 12^{\circ}$ using the Pythagorean identity:

$\cos^2 a + \sin^2 a = 1$

In this case, we have:

$\cos^2 12^{\circ} + \sin^2 12^{\circ} = 1$

Subtracting $\sin^2 12^{\circ}$ from both sides of the equation, we get:

$\cos^2 12^{\circ} = 1 - \sin^2 12^{\circ}$

Substituting this expression into the double-angle formula, we get:

$\cos 24^{\circ} = 2(1 - \sin^2 12^{\circ}) - 1$

Simplifying the expression, we get:

$\cos 24^{\circ} = 1 - 2\sin^2 12^{\circ}$

Therefore, we can express $\sin 114^{\circ}$ in terms of $\sin 12^{\circ}$ as:

$\sin 114^{\circ} = 1 - 2\sin^2 12^{\circ}$

Q: How can we express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$?

A: We can express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ using the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 12^{\circ} = \sin (6^{\circ} + 6^{\circ})$

Using the angle addition formula, we get:

$\sin 12^{\circ} = \sin 6^{\circ} \cos 6^{\circ} + \cos 6^{\circ} \sin 6^{\circ}$

Simplifying the expression, we get:

$\sin 12^{\circ} = 2\sin 6^{\circ} \cos 6^{\circ}$

We can rewrite $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ by dividing both sides of the equation by 2:

$\sin 6^{\circ} \cos 6^{\circ} = \frac{\sin 12^{\circ}}{2}$

Q: How can we express $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$?

A: We can express $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ using the Pythagorean identity:

$\cos^2 a + \sin^2 a = 1$

In this case, we have:

$\cos^2 6^{\circ} + \sin^2 6^{\circ} = 1$

Subtracting $\sin^2 6^{\circ}$ from both sides of the equation, we get:

$\cos^2 6^{\circ} = 1 - \sin^2 6^{\circ}$

We can rewrite $\sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ using the angle addition formula for sine:

$\sin (a + b) = \sin a \cos b + \cos a \sin b$

In this case, we have:

$\sin 12^{\circ} = \sin (6^{\circ} + 6^{\circ})$

Using the angle addition formula, we get:

$\sin 12^{\circ} = \sin 6^{\circ} \cos 6^{\circ} + \cos 6^{\circ} \sin 6^{\circ}$

Simplifying the expression, we get:

$\sin 12^{\circ} = 2\sin 6^{\circ} \cos 6^{\circ}$

We can rewrite $\sin 6^{\circ} \cos 6^{\circ}$ in terms of $\sin 12^{\circ}$ by dividing both sides of the equation by 2:

$\sin 6^{\circ} \cos 6^{\circ} = \frac{\sin 12^{\circ}}{2}$

Substituting this expression into the Pythagorean identity, we get:

$\sin^2 6^{\circ} = \left(\frac{\sin 12^{\circ}}{2}\right)^2$

Simplifying the expression, we get:

$\sin^2 6^{\circ} = \frac{\sin^2 12^{\circ}}{4}$

Substituting this expression into the Pythagorean identity, we get:

$\cos^2 6^{\circ} = 1 - \frac{\sin^2 12^{\circ}}{4}$

We can rewrite $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$ by subtracting $\sin^2 6^{\circ}$ from both sides of the equation:

$\cos^2 6^{\circ} - \sin^2 6^{\circ} = 1 - \frac{\sin^2 12^{\circ}}{4} - \frac{\sin^2 12^{\circ}}{4}$

Simplifying the expression, we get:

$\cos^2 6^{\circ} - \sin^2 6^{\circ} = 1 - \frac{\sin^2 12^{\circ}}{2}$

Conclusion

In this article, we have answered some frequently asked questions related to expressing trigonometric functions in terms of $\sin 12^{\circ}$. We have derived expressions for $\tan 12^{\circ}$, $\sin 114^{\circ}$, $\sin 6^{\circ} \cos 6^{\circ}$, and $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of $\sin 12^{\circ}$. We hope that this article has been helpful in understanding the relationship between these trigonometric functions.