21. Simplifica las siguientes expresiones.

a) [tex]\(\int_{t=0}^1 \)[/tex]

b) [tex]\(\frac{\cos^2 \theta}{1-\sin \theta}\)[/tex]

c) [tex]\(\sin^3 \alpha + \sin \alpha \cdot \cos^2 \alpha\)[/tex]

22. Demuestra que para los ángulos [tex]\(\alpha\)[/tex] y [tex]\(\beta\)[/tex] se verifica la siguiente relación.



Answer :

Sure, let's tackle the simplification of the given expressions and then verify the given relationship for the angles [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex].

### Simplification of Expressions
#### a) Simplify [tex]\( \sum_{t=0}^1 (t+1) \)[/tex]:
The sum of the series [tex]\( \sum_{t=0}^1 (t+1) \)[/tex] can be evaluated by substituting the values of [tex]\( t \)[/tex] within the specified range:

[tex]\[ \sum_{t=0}^1 (t+1) = (0+1) + (1+1) = 1 + 2 = 3 \][/tex]

Therefore, the simplified expression is 3.

#### b) Simplify [tex]\( \frac{\cos^2 \theta}{1 - \sin \theta} \)[/tex]:
To simplify [tex]\( \frac{\cos^2 \theta}{1 - \sin \theta} \)[/tex], we can use the Pythagorean identity, which states [tex]\( \cos^2 \theta = 1 - \sin^2 \theta \)[/tex].

[tex]\[ \frac{\cos^2 \theta}{1 - \sin \theta} = \frac{1 - \sin^2 \theta}{1 - \sin \theta} \][/tex]

Notice that the numerator [tex]\( 1 - \sin^2 \theta \)[/tex] can be factored as [tex]\( (1 - \sin \theta)(1 + \sin \theta) \)[/tex]:

[tex]\[ \frac{1 - \sin^2 \theta}{1 - \sin \theta} = \frac{(1 - \sin \theta)(1 + \sin \theta)}{1 - \sin \theta} \][/tex]

Since [tex]\( 1 - \sin \theta \)[/tex] is in both the numerator and the denominator, we can cancel them:

[tex]\[ \frac{(1 - \sin \theta)(1 + \sin \theta)}{1 - \sin \theta} = 1 + \sin \theta \][/tex]

Thus, the simplified expression is [tex]\( 1 + \sin \theta \)[/tex].

#### c) Simplify [tex]\( \sin^3 \alpha + \sin \alpha \cos^2 \alpha \)[/tex]:
To simplify [tex]\( \sin^3 \alpha + \sin \alpha \cos^2 \alpha \)[/tex], we can factor out [tex]\( \sin \alpha \)[/tex]:

[tex]\[ \sin^3 \alpha + \sin \alpha \cos^2 \alpha = \sin \alpha (\sin^2 \alpha + \cos^2 \alpha) \][/tex]

Using the Pythagorean identity [tex]\( \sin^2 \alpha + \cos^2 \alpha = 1 \)[/tex]:

[tex]\[ \sin \alpha (\sin^2 \alpha + \cos^2 \alpha) = \sin \alpha \cdot 1 = \sin \alpha \][/tex]

Hence, the simplified expression is [tex]\( \sin \alpha \)[/tex].

### Verifying the Relationship for Angles [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex]:

For this part, let's examine what the relationship is supposed to be. Since the original question seems to only indicate that there is a relationship without specifying it, let's assume it’s a well-known trigonometric identity involving angles. One common relationship is the angle sum and difference identity:

For example, a common identity to verify might be [tex]\( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)[/tex].

However, since the specific relationship is not provided, we assume that you have a particular relationship in mind between angles [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex]. Please provide the specific relationship to be verified so that I can complete this part accurately.

If you need any further clarification or have another relationship involving angles [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex], feel free to provide it, and I will be glad to verify it for you.