Question

Sin^2(п-t)/1+sin(3п/2+t)=cos(2п-t)

Answer 1

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Answer 2

Let's simplify the left-hand side of the equation using some trigonometric identities:

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

sin(3п/2+t) = sin(3п/2)cos(t) + cos(3п/2)sin(t) = -cos(t)

Now we can substitute these values into the original equation and simplify:

sin^2(t)/(1 - cos(t)) = cos(2п - t)

Next, we can use the identity cos(2п - t) = cos(t) to simplify further:

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

Multiplying both sides by (1 - cos(t)), we get:

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

Using the identity cos^2(t) + sin^2(t) = 1, we can substitute cos^2(t) = 1 - sin^2(t) to get:

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

Simplifying, we get:

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

Using the identity sin^2(t) + cos^2(t) = 1, we can substitute sin^2(t) = 1 - cos^2(t) to get:

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

Expanding and simplifying, we get:

2cos^2(t) + cos(t) - 3 = 0

This is a quadratic equation in cos(t), which we can solve using the quadratic formula:

cos(t) = (-1 ± sqrt(13))/4

So the solutions for t are:

t = acos((-1 + sqrt(13))/4) and t = acos((-1 - sqrt(13))/4)

Note that we took the inverse cosine function to find the values of t, but we need to make sure that these values satisfy the original equation.