Question

1-cost/sint = tg t/2

Answer 1

Смотри вов ложении.

Answer 2

To solve the equation (1 - cos(t)) / sin(t) = tan(t/2), we can use trigonometric identities to simplify and solve for t.

First, let's manipulate the left-hand side of the equation using the double angle identity for cosine: cos(2θ) = 1 - 2sin^2(θ).

(1 - cos(t)) / sin(t) = tan(t/2)

1 - cos(t) = sin(t) * tan(t/2)

Now, using the identity tan(θ) = sin(θ) / cos(θ), we can replace tan(t/2) with sin(t/2) / cos(t/2):

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

Next, simplify the expression by dividing both sides by cos(t) and multiplying both sides by 2:

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

Now, apply the double-angle identity for sine: sin(2θ) = 2sin(θ)cos(θ). This yields:

2(1 - cos(t)) = sin(t) * sin(t)

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

2(1 - cos(t)) = sin^2(t)

Divide both sides by 2:

1 - cos(t) = sin^2(t) / 2

Now, use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1:

1 - cos(t) = (1 - cos^2(t)) / 2

Multiply both sides by 2:

2 - 2cos(t) = 1 - cos^2(t)

Rearrange the terms:

cos^2(t) - 2cos(t) + 1 = 0

This is a quadratic equation in terms of cos(t). Let's solve it by factoring:

(cos(t) - 1)^2 = 0

Now, take the square root of both sides:

cos(t) - 1 = 0

Add 1 to both sides:

cos(t) = 1

Now, we have found that cos(t) = 1. To find the solutions for t, we can use the inverse cosine function (also known as arccos or cos^-1):

t = cos^-1(1)

The value of t that satisfies the equation is:

t = 0 degrees or t = 2π radians

Remember that trigonometric functions have periodicity, so there are infinite solutions for t, but the main solution within a given interval (0 ≤ t < 2π) is t = 0 degrees or t = 2π radians.