Find the exact values of sin 2 theta, cos 2 theta and tan 2thea using the double angle formula
sin theta = 3/4 , pi/2 < theta < pi
given: sinØ = 3/4, and Ø is in quadrant II
then by Pythagoras, y= 3, x = ? , r = 4
x^2 + 9 = 16
x = ±√5 , but in II x = -√7
so cosØ = -√7/4
sin 2Ø = 2sinØcosØ
= 2(3/4)(-√7/4) = -6√7/16
= -3√7/8
cos 2Ø = cos^2 Ø - sin^2 Ø = 7/16 - 9/16
= -2/16
= - 1/8
tan 2Ø = sin 2Ø / cos 2Ø
= (-3√7/8) / (-1/8)
= 3√7
To find the values of sin 2θ, cos 2θ, and tan 2θ, we'll first use the given value of sin θ and the range of θ to determine the values of sin θ, cos θ, and tan θ. Then we'll use the double angle formula to find their respective double angle values.
Given:
sin θ = 3/4
π/2 < θ < π
First, we need to find cos θ and tan θ using the given value of sin θ.
Using the Pythagorean identity:
cos^2 θ = 1 - sin^2 θ
cos^2 θ = 1 - (3/4)^2
cos^2 θ = 1 - 9/16
cos^2 θ = 16/16 - 9/16
cos^2 θ = 7/16
Since θ is in the range π/2 < θ < π, cos θ is negative.
cos θ = -√(7/16)
cos θ = -√7/4
cos θ = -√7/4
Next, we can find tan θ using the values of sin θ and cos θ:
tan θ = sin θ / cos θ
tan θ = (3/4) / (-√7/4)
tan θ = (3/4) * (-4/√7)
tan θ = -3/√7
Now that we know the values of sin θ, cos θ, and tan θ, we can use the double angle formulas to find sin 2θ, cos 2θ, and tan 2θ.
The double angle formulas are:
sin 2θ = 2sin θ cos θ
cos 2θ = cos^2 θ - sin^2 θ
tan 2θ = 2tan θ / (1 - tan^2 θ)
Let's find sin 2θ first:
sin 2θ = 2sin θ cos θ
sin 2θ = 2(3/4) * (-√7/4)
sin 2θ = -3√7 / 8
Next, let's find cos 2θ:
cos 2θ = cos^2 θ - sin^2 θ
cos 2θ = (-√7/4)^2 - (3/4)^2
cos 2θ = (7/16) - (9/16)
cos 2θ = -2/16
cos 2θ = -1/8
Lastly, let's find tan 2θ:
tan 2θ = 2tan θ / (1 - tan^2 θ)
tan 2θ = 2(-3/√7) / (1 - (-3/√7)^2)
tan 2θ = -6√7 / (1 - 9/7)
tan 2θ = -6√7 / (-2/7)
tan 2θ = 21√7
Therefore, the exact values of sin 2θ, cos 2θ, and tan 2θ are:
sin 2θ = -3√7 / 8
cos 2θ = -1/8
tan 2θ = 21√7