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Need help quickly! Let theta be an acute angle of a right triangle. Find the values of the other five trigonometric functions of theta.

Need help quickly! Let theta be an acute angle of a right triangle. Find the values-example-1

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Explanation:

sec(theta) = 1/cos(theta)

cot(theta) = 1/tan(theta) = cos(theta)/sin(theta)

as

tan(theta) = sin(theta)/cos(theta)

sin²(theta) + cos²(theta) = 1

csc(theta) = 1/sin(theta)

5.

sec(theta) = sqrt(73)/8 = 1/cos(theta)

cos(theta) = 1/sec(theta) = 8/sqrt(73)

sin(theta) = sqrt(1 - (8/sqrt(73))²) = sqrt(1 - 64/73) =

= sqrt(73/73 - 64/73) = sqrt(9/73) = 3/sqrt(73)

tan(theta) = 3/sqrt(73) / 8/sqrt(73) = 3/8

cot(theta) = 1/tan(theta) = 8/3

csc(theta) = 1/sin(theta) = sqrt(73)/3

6.

cot(theta) = sqrt(3) = cos(theta)/sin(theta)

we could now say

cos(theta) = sqrt(3)

sin(theta) = 1

but sqrt(3) > 1. and that cannot be. sine and cosine must both be between -1 and +1.

so, we have to make a detour :

cos(theta)/sin(theta) = sqrt(3)

cos(theta) = sqrt(3)×sin(theta)

sin²(theta) + cos²(theta) = 1

sin²(theta) + (sqrt(3)×sin(theta))² = 1

sin²(theta) + 3×sin²(theta) = 1

4×sin²(theta) = 1

sin²(theta) = 1/4

sin(theta) = sqrt(1/4) = 1/2

cos(theta) = sqrt(3)×1/2 = sqrt(3)/2

tan(theta) = 1/cot(theta) = 1/sqrt(3)

csc(theta) = 1/sin(theta) = 1 / 1/2 = 2

sec(theta) = 1/cos(theta) = 1 / sqrt(3)/2 = 2/sqrt(3)

answered
User Zelter Ady
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