Expressing Trigonometric Functions in Terms of m

Oct 13, 2025 by vinooalluri.com 123 views

Expressing Trigonometric Functions in Terms of m

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

Expressing $\tan 12^{\circ}$ in Terms of m

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

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

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\tan 12^{\circ} = \frac{m}{\cos 12^{\circ}}

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

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

Substituting the given value, $\sin 12^{\circ}=m$ , we get:

m^2 + \cos^2 12^{\circ} = 1

Solving for $\cos 12^{\circ}$ , we get:

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

Substituting this value into the equation for $\tan 12^{\circ}$ , we get:

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

Expressing $\sin 114^{\circ}$ in Terms of m

To express $\sin 114^{\circ}$ in terms of m, we can use the fact that $\sin 114^{\circ} = \sin (180^{\circ} - 66^{\circ})$ .

Using the identity $\sin (180^{\circ} - x) = \sin x$ , we get:

\sin 114^{\circ} = \sin 66^{\circ}

Using the angle addition formula, we can rewrite $\sin 66^{\circ}$ as:

\sin 66^{\circ} = \sin (60^{\circ} + 6^{\circ})

Using the angle addition formula, we get:

\sin 66^{\circ} = \sin 60^{\circ} \cos 6^{\circ} + \cos 60^{\circ} \sin 6^{\circ}

Using the values of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$ , we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sin 6^{\circ}

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sin 6^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{1 - \cos^2 6^{\circ}}

Using the Pythagorean identity, we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{1 - (1 - \sin^2 6^{\circ})}

Simplifying the equation, we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{\sin^2 6^{\circ}}

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{m^2}

Simplifying the equation, we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} m

Expressing $\sin 6^{\circ} \cos 6^{\circ}$ in Terms of m

To express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of m, we can use the fact that $\sin 6^{\circ} \cos 6^{\circ} = \frac{1}{2} \sin 12^{\circ}$ .

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 6^{\circ} \cos 6^{\circ} = \frac{1}{2} m

Expressing $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in Terms of m

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

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

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

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

Expressing $2 \sin^2 6^{\circ} \cos^2 6^{\circ}$ in Terms of m

To express $2 \sin^2 6^{\circ} \cos^2 6^{\circ}$ in terms of m, we can use the fact that $2 \sin^2 6^{\circ} \cos^2 6^{\circ} = \sin^2 12^{\circ}$ .

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

2 \sin^2 6^{\circ} \cos^2 6^{\circ} = m^2

In this article, we have expressed various trigonometric functions in terms of a given value, m, where $\sin 12^{\circ}=m$ . We have used the given value to find the values of $\tan 12^{\circ}$ , $\sin 114^{\circ}$ , $\sin 6^{\circ} \cos 6^{\circ}$ , $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ , and $2 \sin^2 6^{\circ} \cos^2 6^{\circ}$ .
Q&A: Expressing Trigonometric Functions in Terms of m

In our previous article, we explored the relationship between various trigonometric functions and expressed them in terms of a given value, m, where $\sin 12^{\circ}=m$ . In this article, we will answer some frequently asked questions related to expressing trigonometric functions in terms of m.

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

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

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

Q: How do you express $\sin 114^{\circ}$ in terms of m?

A: To express $\sin 114^{\circ}$ in terms of m, we can use the fact that $\sin 114^{\circ} = \sin (180^{\circ} - 66^{\circ})$ .

Using the identity $\sin (180^{\circ} - x) = \sin x$ , we get:

\sin 114^{\circ} = \sin 66^{\circ}

Using the angle addition formula, we can rewrite $\sin 66^{\circ}$ as:

\sin 66^{\circ} = \sin (60^{\circ} + 6^{\circ})

Using the angle addition formula, we get:

\sin 66^{\circ} = \sin 60^{\circ} \cos 6^{\circ} + \cos 60^{\circ} \sin 6^{\circ}

Using the values of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$ , we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sin 6^{\circ}

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sin 6^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{1 - \cos^2 6^{\circ}}

Using the Pythagorean identity, we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{1 - (1 - \sin^2 6^{\circ})}

Simplifying the equation, we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{\sin^2 6^{\circ}}

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} \sqrt{m^2}

Simplifying the equation, we get:

\sin 66^{\circ} = \frac{\sqrt{3}}{2} \cos 6^{\circ} + \frac{1}{2} m

Q: How do you express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of m?

A: To express $\sin 6^{\circ} \cos 6^{\circ}$ in terms of m, we can use the fact that $\sin 6^{\circ} \cos 6^{\circ} = \frac{1}{2} \sin 12^{\circ}$ .

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

\sin 6^{\circ} \cos 6^{\circ} = \frac{1}{2} m

Q: How do you express $\cos^2 6^{\circ} - \sin^2 6^{\circ}$ in terms of m?

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

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

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

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

Q: How do you express $2 \sin^2 6^{\circ} \cos^2 6^{\circ}$ in terms of m?

A: To express $2 \sin^2 6^{\circ} \cos^2 6^{\circ}$ in terms of m, we can use the fact that $2 \sin^2 6^{\circ} \cos^2 6^{\circ} = \sin^2 12^{\circ}$ .

Using the given value, $\sin 12^{\circ}=m$ , we can rewrite the equation as:

2 \sin^2 6^{\circ} \cos^2 6^{\circ} = m^2

In this article, we have answered some frequently asked questions related to expressing trigonometric functions in terms of m. We have used the given value, $\sin 12^{\circ}=m$ , to find the values of various trigonometric functions.