sin theta = 8/17 and theta is in the second quadrant. Find sin(2theta),cos(2theta),tan(2theta) exactly

sin(2theta)

cos(2theta)

tan(2theta)

Did it occur to you to find theta first?

Theta=arcsin8/17 =0.489957326 rad in the first quadrant, or in the second quadrant,
PI- 0.489957326 rad= 2.65163533rad
Put each of these in your google search windown:
sin(2* 2.65163533)
cos(2*2.6516)
etc

To find sin(2θ), cos(2θ), and tan(2θ) when sin(θ) = 8/17 and θ is in the second quadrant, we first need to determine the value of θ.

Given that sin(θ) = 8/17, we can find θ by using the inverse sine function or arcsin.

θ = arcsin(8/17)

Since θ is in the second quadrant, we know that both sine and cosine are positive, while tangent is negative.

sin(2θ) can be found using the double-angle formula for sine:

sin(2θ) = 2*sin(θ)*cos(θ)

cos(2θ) can be found using the double-angle formula for cosine:

cos(2θ) = cos²(θ) - sin²(θ)

tan(2θ) can be found using the double-angle formula for tangent:

tan(2θ) = (2*tan(θ))/(1 - tan²(θ))

Now, let's calculate sin(2θ), cos(2θ), and tan(2θ) step by step:

Step 1: Finding θ
θ = arcsin(8/17)
θ ≈ 30.5 degrees

Step 2: Using θ to find sin(2θ)
sin(2θ) = 2*sin(θ)*cos(θ)
substituting sin(θ) = 8/17 and cos(θ) = √(1 - sin²(θ)), we get:
sin(2θ) = 2*(8/17)*√(1 - (8/17)²)
simplifying the expression, we have:
sin(2θ) = 16/17 √(1 - 64/289)
sin(2θ) = 16√(289 - 64)/17√(289)
sin(2θ) = 16√(225)/17√(17)
sin(2θ) = 16/17

Step 3: Using θ to find cos(2θ)
cos(2θ) = cos²(θ) - sin²(θ)
substituting sin(θ) = 8/17 and cos(θ) = √(1 - sin²(θ)), we get:
cos(2θ) = (√(1 - (8/17)²))² - (8/17)²
cos(2θ) = (√(1 - 64/289))² - (64/289)
cos(2θ) = (√(289 - 64)/17)² - 64/289
cos(2θ) = (√(225)/17)² - 64/289
cos(2θ) = (15/17)² - 64/289
cos(2θ) = 225/289 - 64/289
cos(2θ) = 161/289

Step 4: Using θ to find tan(2θ)
tan(2θ) = (2*tan(θ))/(1 - tan²(θ))
substituting tan(θ) = sin(θ)/cos(θ), we get:
tan(2θ) = (2*(8/17))/(1 - (8/17)²)
tan(2θ) = (16/17)/(1 - 64/289)
tan(2θ) = (16/17)/(289/289 - 64/289)
tan(2θ) = (16/17)/(225/289)
tan(2θ) = (16/17)*(289/225)
tan(2θ) = 16/15

Therefore, when sin(θ) = 8/17 and θ is in the second quadrant:
sin(2θ) = 16/17
cos(2θ) = 161/289
tan(2θ) = 16/15

To find sin(2theta), cos(2theta), and tan(2theta) from sin(theta) = 8/17 in the second quadrant, we need to use the double-angle formulas.

1. Find cos(theta) using the Pythagorean identity: cos(theta) = √(1 - sin^2(theta))
cos(theta) = √(1 - (8/17)^2)
= √(1 - 64/289)
= √(225/289)
= 15/17

2. sin(2theta) = 2sin(theta)cos(theta)
sin(2theta) = 2 * (8/17) * (15/17)
= 240/289

3. cos(2theta) = cos^2(theta) - sin^2(theta)
cos(2theta) = (15/17)^2 - (8/17)^2
= 225/289 - 64/289
= 161/289

4. tan(2theta) = sin(2theta)/cos(2theta)
tan(2theta) = (240/289)/(161/289)
= 240/161
= 24/17

Therefore, sin(2theta) = 240/289, cos(2theta) = 161/289, and tan(2theta) = 24/17.