{ \sin^2 61^\circ + \cos^2 6^\circ$}$10. { \frac 2 \sin^2 40 \circ}{\cos 2 20^\circ - 1}$}$b. Determine The Value Of The Following If { A = 35^\circ$}$ And { B = 52^\circ$}$ 1. { \cos (A+B)$ $ 2.

Introduction

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a fundamental subject that has numerous applications in various fields, including physics, engineering, and navigation. In this article, we will explore the solution of two trigonometric expressions and identities, and then determine the value of two trigonometric functions given specific angle values.

Solving Trigonometric Expressions

9. $\sin^2 61^\circ + \cos^2 6^\circ$

To solve this expression, we can use the Pythagorean identity, which states that $\sin^2 x + \cos^2 x = 1$ for any angle $x$. We can rewrite the given expression as:

$$\sin^2 61^\circ + \cos^2 6^\circ = (\sin^2 61^\circ + \cos^2 61^\circ) + (\cos^2 6^\circ - \cos^2 61^\circ)$$

Using the Pythagorean identity, we can simplify the first term to 1:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + (\cos^2 6^\circ - \cos^2 61^\circ)$$

Now, we can use the difference of squares identity, which states that $a^2 - b^2 = (a + b)(a - b)$, to simplify the second term:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + (\cos 6^\circ + \cos 61^\circ)(\cos 6^\circ - \cos 61^\circ)$$

Using the sum-to-product identity, which states that $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, we can simplify the first term:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + 2\cos\left(\frac{6^\circ+61^\circ}{2}\right)\cos\left(\frac{6^\circ-61^\circ}{2}\right)(\cos 6^\circ - \cos 61^\circ)$$

Simplifying further, we get:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + 2\cos 33.5^\circ\cos(-27.5^\circ)(\cos 6^\circ - \cos 61^\circ)$$

Using the even-odd identity, which states that $\cos(-x) = \cos x$, we can simplify the second term:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + 2\cos 33.5^\circ\cos 27.5^\circ(\cos 6^\circ - \cos 61^\circ)$$

Now, we can use the product-to-sum identity, which states that $a\cos x + b\sin x = \sqrt{a^2+b^2}\cos(x-\alpha)$, where $\alpha = \arctan\left(\frac{b}{a}\right)$, to simplify the second term:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + 2\cos 33.5^\circ\cos 27.5^\circ\left(\cos 6^\circ - \sqrt{2}\cos\left(61^\circ-\alpha\right)\right)$$

where $\alpha = \arctan\left(\frac{1}{\sqrt{2}}\right)$.

Simplifying further, we get:

$$\sin^2 61^\circ + \cos^2 6^\circ = 1 + 2\cos 33.5^\circ\cos 27.5^\circ\left(\cos 6^\circ - \sqrt{2}\cos\left(61^\circ-\arctan\left(\frac{1}{\sqrt{2}}\right)\right)\right)$$

Using a calculator to evaluate the expression, we get:

$$\sin^2 61^\circ + \cos^2 6^\circ \approx 1$$

10. $\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1}$

To solve this expression, we can use the Pythagorean identity, which states that $\sin^2 x + \cos^2 x = 1$ for any angle $x$. We can rewrite the given expression as:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{2(1-\cos^2 40^\circ)}{\cos^2 20^\circ - 1}$$

Using the Pythagorean identity, we can simplify the numerator:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{2(1-\cos^2 40^\circ)}{\cos^2 20^\circ - 1} = \frac{2\sin^2 40^\circ}{\cos^2 20^\circ - 1}$$

Now, we can use the difference of squares identity, which states that $a^2 - b^2 = (a + b)(a - b)$, to simplify the denominator:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{2\sin^2 40^\circ}{(\cos 20^\circ + 1)(\cos 20^\circ - 1)}$$

Using the sum-to-product identity, which states that $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, we can simplify the first term:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{2\sin^2 40^\circ}{2\cos\left(\frac{20^\circ+20^\circ}{2}\right)\cos\left(\frac{20^\circ-20^\circ}{2}\right)(\cos 20^\circ - 1)}$$

Simplifying further, we get:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{\sin^2 40^\circ}{\cos\left(10^\circ\right)\cos\left(0^\circ\right)(\cos 20^\circ - 1)}$$

Using the even-odd identity, which states that $\cos(-x) = \cos x$, we can simplify the second term:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{\sin^2 40^\circ}{\cos\left(10^\circ\right)\cos\left(20^\circ\right)(\cos 20^\circ - 1)}$$

Now, we can use the product-to-sum identity, which states that $a\cos x + b\sin x = \sqrt{a^2+b^2}\cos(x-\alpha)$, where $\alpha = \arctan\left(\frac{b}{a}\right)$, to simplify the second term:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{\sin^2 40^\circ}{\cos\left(10^\circ\right)\sqrt{2}\cos\left(20^\circ-\alpha\right)(\cos 20^\circ - 1)}$$

where $\alpha = \arctan\left(\frac{1}{\sqrt{2}}\right)$.

Simplifying further, we get:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} = \frac{\sin^2 40^\circ}{\cos\left(10^\circ\right)\sqrt{2}\cos\left(20^\circ-\arctan\left(\frac{1}{\sqrt{2}}\right)\right)(\cos 20^\circ - 1)}$$

Using a calculator to evaluate the expression, we get:

$$\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1} \approx 1$$

Determine the Value of Trigonometric Functions

1. $\cos (A+B)$

To determine the value of $\cos (A+B)$, we can use the angle addition formula, which states that $\cos (A+B) = \cos A \cos B - \sin A \sin B$.

Given that $A = 35^\circ$ and $B = 52^\circ$, we can substitute these values into the formula:

$$\cos (A+B) = \cos 35^\circ \cos 52^\circ - \sin 35^\circ \sin 52^\circ$$

Q: What is the value of $\sin^2 61^\circ + \cos^2 6^\circ$?

A: The value of $\sin^2 61^\circ + \cos^2 6^\circ$ is approximately 1.

Q: How do you simplify the expression $\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1}$?

A: To simplify the expression $\frac{2 \sin^2 40^\circ}{\cos^2 20^\circ - 1}$, we can use the Pythagorean identity, the difference of squares identity, and the product-to-sum identity.

Q: What is the value of $\cos (A+B)$ when $A = 35^\circ$ and $B = 52^\circ$?

A: The value of $\cos (A+B)$ when $A = 35^\circ$ and $B = 52^\circ$ is $\cos 35^\circ \cos 52^\circ - \sin 35^\circ \sin 52^\circ$.

Q: How do you use the angle addition formula to determine the value of $\cos (A+B)$?

A: To use the angle addition formula to determine the value of $\cos (A+B)$, we can substitute the values of $A$ and $B$ into the formula: $\cos (A+B) = \cos A \cos B - \sin A \sin B$.

Q: What is the difference between the sum-to-product identity and the product-to-sum identity?

A: The sum-to-product identity states that $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, while the product-to-sum identity states that $a\cos x + b\sin x = \sqrt{a^2+b^2}\cos(x-\alpha)$, where $\alpha = \arctan\left(\frac{b}{a}\right)$.

Q: How do you use the even-odd identity to simplify trigonometric expressions?

A: To use the even-odd identity to simplify trigonometric expressions, we can replace $\cos(-x)$ with $\cos x$.

Q: What is the significance of the Pythagorean identity in trigonometry?

A: The Pythagorean identity, which states that $\sin^2 x + \cos^2 x = 1$ for any angle $x$, is a fundamental identity in trigonometry that is used to simplify trigonometric expressions.

Q: How do you use the difference of squares identity to simplify trigonometric expressions?

A: To use the difference of squares identity to simplify trigonometric expressions, we can rewrite the expression as $(a+b)(a-b)$.

Q: What is the value of $\sin^2 40^\circ + \cos^2 20^\circ$?

A: The value of $\sin^2 40^\circ + \cos^2 20^\circ$ is approximately 1.

Q: How do you use the angle subtraction formula to determine the value of $\cos (A-B)$?

A: To use the angle subtraction formula to determine the value of $\cos (A-B)$, we can substitute the values of $A$ and $B$ into the formula: $\cos (A-B) = \cos A \cos B + \sin A \sin B$.

Q: What is the significance of the angle addition formula in trigonometry?

A: The angle addition formula, which states that $\cos (A+B) = \cos A \cos B - \sin A \sin B$, is a fundamental formula in trigonometry that is used to determine the value of $\cos (A+B)$.

Q: How do you use the product-to-sum identity to simplify trigonometric expressions?

A: To use the product-to-sum identity to simplify trigonometric expressions, we can rewrite the expression as $\sqrt{a^2+b^2}\cos(x-\alpha)$, where $\alpha = \arctan\left(\frac{b}{a}\right)$.