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?
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.
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?