IDNLearn.com offers a user-friendly platform for finding and sharing knowledge. Find the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
Alright, let's delve into and prove the identities provided.
### Identity 1: [tex]\((\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2\)[/tex]
First, let's expand both terms individually.
#### Expanding [tex]\((\cos \theta + \sin \theta)^2\)[/tex]
[tex]\[ (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]
#### Expanding [tex]\((\cos \theta - \sin \theta)^2\)[/tex]
[tex]\[ (\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]
#### Adding the expanded forms together:
[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 = (\cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta) + (\cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta) \][/tex]
Combine the terms:
[tex]\[ = \cos^2 \theta + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \][/tex]
Simplify:
[tex]\[ = 2 \cos^2 \theta + 2 \sin^2 \theta \][/tex]
Using the Pythagorean identity:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
So we have:
[tex]\[ 2(\cos^2 \theta + \sin^2 \theta) = 2 \cdot 1 = 2 \][/tex]
Thus, the first identity is proven:
[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2 \][/tex]
### Identity 2: [tex]\(\frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta\)[/tex]
First, let's express the right-hand side in terms of cosine and sine functions.
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
So,
[tex]\[ \sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \][/tex]
Combine the fractions:
[tex]\[ = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
Now, let's focus on the left-hand side:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]
To simplify this to match the right-hand side, we can use an algebraic trick by multiplying the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \cdot \frac{1 - \sin \theta}{1 - \sin \theta} \][/tex]
This simplifies to:
[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta) (1 - \sin \theta)} \][/tex]
Using the difference of squares:
[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta \][/tex]
Since [tex]\(1 - \sin^2 \theta = \cos^2 \theta\)[/tex], we get:
[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
We can see this matches the right-hand side:
[tex]\[ = \sec \theta - \tan \theta \][/tex]
Thus, the second identity is proven:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta \][/tex]
In conclusion, we've proven both identities:
1. [tex]\((\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2\)[/tex]
2. [tex]\(\frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta\)[/tex]
### Identity 1: [tex]\((\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2\)[/tex]
First, let's expand both terms individually.
#### Expanding [tex]\((\cos \theta + \sin \theta)^2\)[/tex]
[tex]\[ (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]
#### Expanding [tex]\((\cos \theta - \sin \theta)^2\)[/tex]
[tex]\[ (\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta \][/tex]
#### Adding the expanded forms together:
[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 = (\cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta) + (\cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta) \][/tex]
Combine the terms:
[tex]\[ = \cos^2 \theta + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \][/tex]
Simplify:
[tex]\[ = 2 \cos^2 \theta + 2 \sin^2 \theta \][/tex]
Using the Pythagorean identity:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
So we have:
[tex]\[ 2(\cos^2 \theta + \sin^2 \theta) = 2 \cdot 1 = 2 \][/tex]
Thus, the first identity is proven:
[tex]\[ (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2 \][/tex]
### Identity 2: [tex]\(\frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta\)[/tex]
First, let's express the right-hand side in terms of cosine and sine functions.
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
So,
[tex]\[ \sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \][/tex]
Combine the fractions:
[tex]\[ = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
Now, let's focus on the left-hand side:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]
To simplify this to match the right-hand side, we can use an algebraic trick by multiplying the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \cdot \frac{1 - \sin \theta}{1 - \sin \theta} \][/tex]
This simplifies to:
[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta) (1 - \sin \theta)} \][/tex]
Using the difference of squares:
[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta \][/tex]
Since [tex]\(1 - \sin^2 \theta = \cos^2 \theta\)[/tex], we get:
[tex]\[ = \frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
We can see this matches the right-hand side:
[tex]\[ = \sec \theta - \tan \theta \][/tex]
Thus, the second identity is proven:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta \][/tex]
In conclusion, we've proven both identities:
1. [tex]\((\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta )^2 = 2\)[/tex]
2. [tex]\(\frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta\)[/tex]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
