# Algebra 2 Help: Pythagoras, Trigonometry, and Quadrants?

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

tan Θ

tan2 Θ

1

−1

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

False

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

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).

1

cos2 x

sin2 x

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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?

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

• 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

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