Fiding the exact value on sum and difference identities?

• February 9th 2011, 12:14 PM
homeylova223
Fiding the exact value on sum and difference identities?
Can anyone help me solve these two problem as I am confused on how to them?

1. Find the exact value
cos(x+y) if tan x=5/3 and sin y= 1/3

2.sec(x-y) if csc x=5/3 and tan y=5/12

Can anyone show me how to solve these problems what I am having trouble understanding is what to do with the tan.

For example on number 1 I know you can use sin y to find cos y using square root 1-sin^2 then use cos a cos b+ sin a sin b

But I am confused can anyone show me how to work one of these problems out please?(Crying)
• February 9th 2011, 12:23 PM
skeeter
Quote:

Originally Posted by homeylova223
Can anyone help me solve these two problem as I am confused on how to them?

1. Find the exact value
cos(x+y) if tan x=5/3 and sin y= 1/3

2.sec(x-y) if csc x=5/3 and tan y=5/12

same quadrant info for angles x and y needed here, also.

...
• February 9th 2011, 04:17 PM
homeylova223
Both x and y use quadrant one
• February 9th 2011, 04:29 PM
Keep in mind that this solution is for when it's in Quadrant I, else you will have different signs for the trigonometric ratios.

For #1, we have the identity: cos(x+y) = cos x * cosy - sin x * sin y
so all we have to find are these individual components.

For angle x, we are given tan(x) = 5/3. Draw out the triangle. tan x = O/A (opposite over adjacent)

Find ? using Pythagoras. we then know that cos(x) = A/H (adjacent over hypotenuse). ie. 3/?
and that sin(x) = O/H (opposite over hypotenuse) ie. 5/?

Repeat this process for the other piece of information you have for sin y = 1/3, a diagram with y as your angle and given side length embedded in the ratio. hope that helps.
• February 9th 2011, 04:33 PM
skeeter
$\cos(x+y) = \cos{x}\cos{y} - \sin{x}\sin{y}$

sketch a couple of reference triangles in quad I for angles x and y

$\tan{x} = \dfrac{5}{3} = \dfrac{opp}{adj}$

$\cos{x} = \dfrac{3}{\sqrt{34}} , \sin{x} = \dfrac{5}{\sqrt{34}}
$

$\sin{y} = \dfrac{1}{3} = \dfrac{opp}{hyp}
$

$\cos{y} = \dfrac{\sqrt{8}}{3}$

sub the necessary values into the sum identity and evaluate.
• February 13th 2011, 12:27 PM
homeylova223
I have a quick question I did cos(x+y)= cos (3/square root 34) cos (square root 8/3)- sin (5/square root 34) sin (1/3)
I end up getting .999 but the answer is 12 square root 17- 5 square root 34 / 102. I am not sure what I did to get wrong answer.
• February 13th 2011, 12:50 PM
skeeter
Quote:

Originally Posted by homeylova223
I have a quick question I did cos(x+y)= cos (3/square root 34) cos (square root 8/3) - sin (5/square root 34) sin (1/3) no, no, - no, no

it's $\cos{x} = \dfrac{3}{\sqrt{34}}$ , not $\cos\left( \dfrac{3}{\sqrt{34}}\right)$

$\cos(x+y) = \dfrac{3}{\sqrt{34}} \cdot \dfrac{\sqrt{8}}{3} - \dfrac{5}{\sqrt{34}} \cdot \dfrac{1}{3}$
• February 13th 2011, 02:26 PM
homeylova223
I have one more question? In my second problem I have sec(x-y) if csc x=5/3 and tan y=5/12

So figuring out csc=5/3 tany=12/5 secy=13/5 cscy= 13/12 secx=5/4

I did (5/4)(13/5)+(5/3)(13/12) and I got 5.23 but the correct answer is 65/56 can anyone help me figure out what I did wrong?
• February 13th 2011, 03:34 PM
skeeter
Quote:

Originally Posted by homeylova223
I have one more question? In my second problem I have sec(x-y) if csc x=5/3 and tan y=5/12

So figuring out csc=5/3 tany=12/5 secy=13/5 cscy= 13/12 secx=5/4

I did (5/4)(13/5)+(5/3)(13/12) and I got 5.23 but the correct answer is 65/56 can anyone help me figure out what I did wrong?

$\sec(x-y) = \dfrac{1}{\cos(x-y)} = \dfrac{1}{\cos{x}\cos{y} + \sin{x}\sin{y}}$

$\csc{x} = \dfrac{5}{3}$ , $\sin{x} = \dfrac{3}{5}$ , $\cos{x} = \dfrac{4}{5}$

$\csc{y} = \dfrac{13}{12}$ , $\sin{x} = \dfrac{12}{13}$ , $\cos{x} = \dfrac{5}{13}$