Question

Cos t=-5/13 8.5п<t<9п вычислить sin(-t)

Answer 1

Sin^2(A)+cos^2(A)=1
sin^2(A)=1-cos^2(A)
sina=(1-cos^2(A))^(1/2)

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Answer 2

To calculate sin(-t), we need to determine the value of -t within the given interval. Since the interval is 8.5π < t < 9π and cos(t) = -5/13, we can find the value of t as follows:

cos(t) = -5/13

Using the identity sin^2(t) + cos^2(t) = 1, we can find sin(t):

sin^2(t) = 1 - cos^2(t) sin^2(t) = 1 - (-5/13)^2 sin^2(t) = 1 - 25/169 sin^2(t) = (169 - 25) / 169 sin^2(t) = 144/169

Taking the square root of both sides, we get:

sin(t) = ±sqrt(144/169)

Now, since we want to find sin(-t), we can substitute -t into the equation:

sin(-t) = ±sqrt(144/169)

Note that both sin(t) and sin(-t) have the same absolute value, so sin(-t) = ±sqrt(144/169). The sign will depend on the quadrant in which t lies.

Since 8.5π < t < 9π, we can conclude that t is in the fourth quadrant where both sine and cosine are negative. Therefore, sin(-t) will also be negative.

Hence, sin(-t) = -sqrt(144/169) or approximately -12/13.