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Verify the following trigonometric identity:

[tex]\[
(1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A)
\][/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's solve this trigonometric equation step-by-step. The given equation is:

[tex]\[ (1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A) \][/tex]

To solve this equation, we need to simplify and verify both sides separately.

### Step 1: Simplify the Left-Hand Side (LHS)
The left-hand side of the equation is:

[tex]\[ (1 - \sin A + \cos A)^2 \][/tex]

We'll expand this expression using the binomial theorem:

[tex]\[ (1 - \sin A + \cos A)^2 = (1 - \sin A + \cos A)(1 - \sin A + \cos A) \][/tex]

Using the distributive property (also known as FOIL for binomials), we get:

[tex]\[ (1 - \sin A + \cos A)(1 - \sin A + \cos A) = 1 \cdot 1 + 1 \cdot (-\sin A) + 1 \cdot \cos A - \sin A \cdot 1 - \sin A \cdot (-\sin A) - \sin A \cdot \cos A + \cos A \cdot 1 + \cos A \cdot (-\sin A) + \cos A \cdot \cos A \][/tex]

Combining like terms:

[tex]\[ = 1 - \sin A + \cos A - \sin A + \sin^2 A - \sin A \cos A + \cos A - \sin A \cos A + \cos^2 A \][/tex]

[tex]\[ = 1 - 2\sin A + \cos A + \cos A + \sin^2 A + \cos^2 A - 2 \sin A \cos A \][/tex]

Since [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]:

[tex]\[ = 1 - 2 \sin A + 2 \cos A + 1 - 2 \sin A \cos A \][/tex]

[tex]\[ = 2 - 2 \sin A + 2 \cos A - 2 \sin A \cos A \][/tex]

To simplify further:

[tex]\[ = 2 + 2 (\cos A - \sin A - \sin A \cos A) \][/tex]

This is already simplified.

### Step 2: Simplify the Right-Hand Side (RHS)
The right-hand side of the equation is:

[tex]\[ 2(1 - \sin A)(1 + \cos A) \][/tex]

We'll expand this expression:

[tex]\[ 2(1 - \sin A)(1 + \cos A) = 2 \times [(1 \cdot 1) + (1 \cdot \cos A) - (\sin A \cdot 1) - (\sin A \cdot \cos A)] \][/tex]

Combining like terms:

[tex]\[ = 2(1 + \cos A - \sin A - \sin A \cos A) \][/tex]

Distribute the 2:

[tex]\[ = 2 \cdot 1 + 2 \cos A - 2 \sin A - 2 \sin A \cos A \][/tex]

[tex]\[ = 2 + 2 \cos A - 2 \sin A - 2 \sin A \cos A \][/tex]

### Step 3: Comparing LHS and RHS
Now, both sides have been simplified:

[tex]\[ \text{LHS}: 2 + 2 \cos A - 2 \sin A - 2 \sin A \cos A \][/tex]

[tex]\[ \text{RHS}: 2 + 2 \cos A - 2 \sin A - 2 \sin A \cos A \][/tex]

Interestingly, the simplified forms of the left-hand side and right-hand side match exactly. Thus, the equation:

[tex]\[ (1 - \sin A + \cos A)^2 = 2(1 - \sin A)(1 + \cos A) \][/tex]

is indeed true, as the simplified expressions of both sides are equivalent.
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