if cos(θ)=10/26, 0≤θ≤π/2 determine the following. Leave results as exact values.
sin(θ)=
tan(θ)=
sec(θ)=
my bad, I read the domain as
0 ≤ θ ≤ 2π , need to clean my glasses
sinθ = 12/13
tanθ = 6/5
secθ = 1/cosθ = 13/5
First simplify value of cos θ
cos θ = 10 / 26 = 2 ∙ 5 / 2 ∙ 13
cos θ = 5 / 13
sin θ = ± √ ( 1 - cos² θ )
0 ≤ θ ≤ π / 2 is Quadrant I
In Quadrant I all trigonometric functions are positive so:
sin θ = √ ( 1 - cos² θ )
sin θ = √ [ 1 - ( 5 / 13 )² ]
sin θ = √ ( 1 - 25 / 169 )
sin θ = √ ( 169 / 169 - 25 / 169 )
sin θ = √ ( 144 / 169 )
sin θ = √144 / √169
sin θ = 12 / 13
tan θ = sin θ / cos θ = ( 12 / 13 ) / ( 5 / 13 ) = 13 ∙ 12 / 5 ∙ 13 = 12 / 5
sec θ = 1 / cos θ = 1 / ( 5 / 13 ) = 13 / 5
Draw the triangle in QI. It is your standard 5-12-13 triangle, scaled up, with
x = 10
r = 26
y = 24
sinθ = y/r
tanθ = y/x
secθ = 1/cosθ
sketch your triangle,
cosθ = 10/23 = 5/13 , since the cosine is + in I or IV, θ is in I or IV
either recognize the 5-12-13 triangle or use Pythagoras
x = 5, y= 6, r = 13
sinθ = 12/13 or sinθ = -12/13
tanθ = 6/5 or tanθ = -6/5
secθ = 1/cosθ = 13/5
Oh, I see you want to have some fun with trigonometry! Great! Let's solve this problem and sprinkle some humor along the way.
To find the value of sin(θ), we can use the Pythagorean identity sin²(θ) + cos²(θ) = 1. So, substituting the given value of cos(θ), we have:
sin²(θ) + (10/26)² = 1.
Let's simplify it a bit:
sin²(θ) + 100/676 = 1.
Subtracting 100/676 from both sides:
sin²(θ) = 576/676.
Now, let's find the square root of both sides:
sin(θ) = ±√(576/676).
Simplifying under the radical:
sin(θ) = ±(24/26).
But since the angle θ lies in the first quadrant (0 ≤ θ ≤ π/2), the value of sin(θ) must be positive:
sin(θ) = 24/26.
Next, let's determine tan(θ). We know that tan(θ) = sin(θ)/cos(θ), so:
tan(θ) = (24/26)/(10/26).
Dividing numerator and denominator by 26:
tan(θ) = 24/10.
And reducing the fraction:
tan(θ) = 12/5.
Finally, let's calculate sec(θ). We know that sec(θ) = 1/cos(θ), so:
sec(θ) = 1/(10/26).
Inverting the fraction:
sec(θ) = 26/10.
Simplifying the fraction:
sec(θ) = 13/5.
So, to recap:
sin(θ) = 24/26,
tan(θ) = 12/5,
sec(θ) = 13/5.
I hope that helps! Remember, laughter is the best cosine, I mean, medicine!
To find the values of sin(θ), tan(θ), and sec(θ) when cos(θ) is given, we can use the fundamental trigonometric identity:
sin^2(θ) + cos^2(θ) = 1
Given that cos(θ) = 10/26, we can substitute this value into the identity:
sin^2(θ) + (10/26)^2 = 1
To solve for sin(θ), we need to find the square root of the remaining term after isolating it:
sin^2(θ) = 1 - (10/26)^2
sin^2(θ) = 1 - 100/676
sin^2(θ) = (676 - 100) / 676
sin^2(θ) = 576 / 676
Taking the square root of both sides, we find:
sin(θ) = √(576 / 676)
sin(θ) = √(576) / √(676)
sin(θ) = 24 / 26
sin(θ) = 12 / 13
So, sin(θ) = 12/13.
Now, we can find tan(θ) and sec(θ) using the given values of sin(θ) and cos(θ):
tan(θ) = sin(θ) / cos(θ)
tan(θ) = (12/13) / (10/26)
tan(θ) = (12/13) * (26/10)
tan(θ) = 312 / 130
tan(θ) = 156/65
So, tan(θ) = 156/65.
sec(θ) = 1 / cos(θ)
sec(θ) = 1 / (10/26)
sec(θ) = 26/10
sec(θ) = 13/5
So, sec(θ) = 13/5.