Algebra 2 Help: Pythagoras, Trigonometry, and Quadrants?

2 Answers

• 1) Simplify sine theta over square root of the quantity 1 minus sine squared theta.

  sin(x) / sqrt( 1- sin^2(x) ) = sin(x) / sqrt( cos^2(x) ) = sin(x) / cos(x) = tan Θ

  2) Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?

  It is true. The square root leads to a positive number. So the answer is in the quadrants where sine is positive. So (B)

  False

  True; quadrants I & IV

  True; quadrants II & III

  True; quadrants I & III

  3) If sin Θ = 5 over 6, what are the values of cos Θ and tan Θ?

  cos(x) = +/- sqrt( 1 – sin^2(x) ) = +/- sqrt( 1 – 25/36 ) =+/- sqrt( 11/36 ) = ± sqt(11) / 6

  tan(x) = sin(x) / cos(x) = +/- 5 / sqrt(11)

  cos Θ = ±11 over 36; tan Θ = ±1 over 11

  cos Θ = ±square root 11 over 6; tan Θ = negative 1 over 11

  cos Θ = ±11 over 6; tan Θ = negative 5 over 11

  cos Θ = ±square root 11 over 6; tan Θ = ±5 square root 11 over 11

  4) Simplify (1 − sin x)(1 + sin x). This is (a + b) (a – b) = a^2 – b^2

  So this is 1 – cos^2(x) = sin^2(x)

  1

  cos2 x

  sin2 x

  tan2 x

  5) If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

  If 3 pi over 2 < Θ < 2π Then the signs of sinx and tanx are both negative.

  cos(x) = sqrt(2)/2. So sin(x) = -sqrt(2)/2 and tan(x) = -1.

  sin Θ = square root 2 over 2; tan Θ = −1

  sin Θ = negative square root 2 over 2; tan Θ = 1

  sin Θ = square root 2 over 2; tan Θ = negative square root 2

  sin Θ = negative square root 2 over 2; tan Θ = −1

• I got them all right