Question

Cos альфа +sin альфа *sin(180°- альфа)＝1

Answer 1

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Answer 2

To prove the trigonometric identity cos(α) + sin(α) * sin(180° - α) = 1, we'll use some trigonometric identities.

1. The sum-to-product identity: sin(A) * sin(B) = (1/2) * [cos(A - B) - cos(A + B)]
2. The angle subtraction identity: cos(A - B) = cos(A) * cos(B) + sin(A) * sin(B)
3. The angle addition identity: cos(A + B) = cos(A) * cos(B) - sin(A) * sin(B)

Now, let's start with the left-hand side of the equation:

LHS = cos(α) + sin(α) * sin(180° - α)

We know that sin(180° - α) = sin(α), so we can rewrite the equation as:

LHS = cos(α) + sin(α) * sin(α)

Now, using the sum-to-product identity with A = B = α:

sin(α) * sin(α) = (1/2) * [cos(α - α) - cos(α + α)]

sin(α) * sin(α) = (1/2) * [cos(0°) - cos(2α)]

Since cos(0°) = 1, the equation becomes:

sin(α) * sin(α) = (1/2) * (1 - cos(2α))

Now, we can rewrite the LHS:

LHS = cos(α) + (1/2) * (1 - cos(2α))

Now, let's focus on the right-hand side of the equation:

RHS = 1

Now, to prove the identity, we need to show that LHS = RHS. So, let's simplify LHS:

LHS = cos(α) + (1/2) * (1 - cos(2α))

We know the double-angle identity: cos(2α) = 2 * cos^2(α) - 1

Now, replace cos(2α) with 2 * cos^2(α) - 1 in the LHS:

LHS = cos(α) + (1/2) * (1 - (2 * cos^2(α) - 1))

LHS = cos(α) + (1/2) * (2 - 2 * cos^2(α) + 1)

LHS = cos(α) + 1 - cos^2(α)

Finally, use the Pythagorean identity: sin^2(α) + cos^2(α) = 1

So, cos^2(α) = 1 - sin^2(α)

Replace cos^2(α) with 1 - sin^2(α) in the LHS:

LHS = cos(α) + 1 - (1 - sin^2(α))

LHS = cos(α) + 1 - 1 + sin^2(α)

LHS = cos(α) + sin^2(α)

Now, use the trigonometric identity: cos^2(α) + sin^2(α) = 1

LHS = 1

Thus, LHS = RHS, and we have proven the identity:

cos(α) + sin(α) * sin(180° - α) = 1