evaluate the expression under the given conditions.
cos(θ − ϕ); cos(θ) = 5/13, θ in Quadrant IV, tan(ϕ) = − square root15,ϕ in Quadrant II
Anyone know how to do this I have never been this confused in my life
start with : cos(θ) = 5/13, θ in Quadrant IV
you should recognize the 5-12-13 right-angled triangle
and since cosØ = adjacent/hypotenuse
x = 5, r = 13 , y = -12, since Ø is in IV
and sinØ = -12/13
also tan(ϕ) = −√15 = -√15/1 = y/x and ϕ is in II,
y = √15 , x = -1
r^2 = x^2 + y^2 = 15+1 = 16
r = 4
sinϕ = √15/4 , cosϕ = -1/4
you must know that:
cos(θ − ϕ) = cosθcosϕ + sinθsinϕ
= (5/13)(-1/4) + (-12/13)(√15/4)
= -5/52 - 12√15/52
= (-5-12√15)/52
To do these type of problems you must know your basic trig definitions as well as the CAST rule.
Well, well, well... it seems like you've stumbled upon a tricky math problem. Don't worry, I'm here to bring some humor to lighten the mood!
Let's break it down, shall we? We know that cos(θ) = 5/13 and θ is in Quadrant IV. Since cos(θ) = adjacent/hypotenuse, we can assign a value of 5 to the adjacent side and 13 to the hypotenuse. The opposite side could be calculated using the ever-handy Pythagorean theorem, but let's leave that job to the mathematicians.
Now, for tan(ϕ) = -√15 with ϕ in Quadrant II. Since tan(ϕ) = opposite/adjacent, we can see that the opposite side is -√15 (negative because we're in Quadrant II) and the adjacent side is 1.
Now, let's go back to our original expression: cos(θ - ϕ). Since we have the values for cos(θ) and cos(ϕ), we can plug them into the expression. So it becomes cos(θ - ϕ) = cos(θ) * cos(ϕ) + sin(θ) * sin(ϕ).
But hold on a second! We don't have the values for sin(θ) and sin(ϕ). Sigh, seems like this problem just got a little more complicated. But hey, don't worry! We've made it this far, and I'm sure you'll figure it out.
So, take a deep breath, consult your trusty math books or resources, and keep on going! Math may be confusing at times, but remember, laughter is the best equation solver. Good luck!
To evaluate the expression cos(θ - ϕ), we need to find the values of cos(θ) and tan(ϕ) first.
Given that cos(θ) = 5/13 and θ is in Quadrant IV, we can use the Pythagorean Identity to find sin(θ):
sin^2(θ) + cos^2(θ) = 1
sin^2(θ) + (5/13)^2 = 1
sin^2(θ) + 25/169 = 1
sin^2(θ) = 144/169
sin(θ) = sqrt(144/169)
sin(θ) = 12/13
Now, given that tan(ϕ) = -sqrt(15) and ϕ is in Quadrant II, we can use the Pythagorean Identity to find cos(ϕ):
tan(ϕ) = sin(ϕ)/cos(ϕ)
-sqrt(15) = (sqrt(15))/cos(ϕ)
cos(ϕ) = -1/sqrt(15)
cos(ϕ) = -sqrt(15)/15
Finally, we can evaluate cos(θ - ϕ) using the difference formula for cosine:
cos(θ - ϕ) = cos(θ)cos(ϕ) + sin(θ)sin(ϕ)
cos(θ - ϕ) = (5/13)(-sqrt(15)/15) + (12/13)(-sqrt(15))
cos(θ - ϕ) = (-5sqrt(15) - 12sqrt(15))/195
cos(θ - ϕ) = -17sqrt(15)/195
Therefore, the value of cos(θ - ϕ) under the given conditions is -17sqrt(15)/195.
To evaluate the expression cos(θ - ϕ) when cos(θ) = 5/13 and tan(ϕ) = -sqrt(15), we need to use the trigonometric identities and Pythagorean identities to find the values of sin(θ), sin(ϕ), cos(θ), cos(ϕ), sin(θ - ϕ), and cos(θ - ϕ).
Given Information:
cos(θ) = 5/13 (θ in Quadrant IV)
tan(ϕ) = -sqrt(15) (ϕ in Quadrant II)
Using the given information, we can determine the values of sin(θ), cos(θ), sin(ϕ), and cos(ϕ).
First, we can find sin(θ) using the Pythagorean identity:
sin^2(θ) + cos^2(θ) = 1
sin^2(θ) = 1 - cos^2(θ)
sin^2(θ) = 1 - (5/13)^2
sin^2(θ) = 1 - 25/169
sin^2(θ) = (169 - 25)/169
sin^2(θ) = 144/169
sin(θ) = sqrt(144/169) = 12/13 (positive value since θ is in Quadrant IV)
Next, we can find cos(ϕ) using the Pythagorean identity:
sin^2(ϕ) + cos^2(ϕ) = 1
sin^2(ϕ) = 1 - cos^2(ϕ)
sin^2(ϕ) = 1 - (1 + tan^2(ϕ))
sin^2(ϕ) = 1 - (1 + (-sqrt(15))^2)
sin^2(ϕ) = 1 - (1 + 15)
sin^2(ϕ) = 1 - 16
sin^2(ϕ) = -15 (There is no real solution for sin(ϕ) since ϕ is in Quadrant II)
Using the known values, we can now find sin(θ - ϕ), cos(θ - ϕ).
sin(θ - ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ)
cos(θ - ϕ) = cos(θ)cos(ϕ) + sin(θ)sin(ϕ)
Plugging in the values we have:
sin(θ - ϕ) = (12/13)(cos(ϕ)) + (5/13)(sin(ϕ))
Since we couldn't find the value of sin(ϕ), we won't be able to find sin(θ - ϕ) or cos(θ - ϕ).