Find sin(α) and cos(β), tan(α) and cot(β), and sec(α) and csc(β). The hypotenuse is 7 and the side is 4.

Solution:

The problem statement can be represented as:

Since it is a right angled triangle we can write

AC² = AB² + BC²

(7)² = (AB)² + (4)²

(AB)² = 49 - 16 = 33

AB = √33

sin(α) = AB/AC = √33/7

cos(β) = AB/AC = √33/7

tan(α) = AB/BC = √33/4

cot(β) = 1/tan(β)

tan(β) = BC/AB = 4/√33

therefore

cot(β) = √33/4

sec(α) = 1/cos(α)

cos(α) = 4/7

Therefore

sec(α) = 1/(4/7) = 7/4

csc(β) = 1/sin(β)

sin(β) = 4/7

Therefore

csc(β) = 1/sin(β) = 1/(4/7)

csc(β) = 7/4

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Find sin(α) and cos(β), tan(α) and cot(β), and sec(α) and csc(β). The hypotenuse is 7 and the side is 4

Summary:

sin(α) = √33/7 and cos(β) = √33/7; tan(α) = √33/4 and cot(β)= √33/4 , and sec(α) = 7/4 and csc(β) = 7/4