# Given: `cosx = (-3)/5 ` , `tany = 1/2` where x is in Quadrant II and 180<y<270degreesCalculate cos (x+y)  (without using a calculator). cot (30degrees + x)

durbanville | High School Teacher | (Level 2) Educator Emeritus

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There is a triangle in Quadrant II  (where x= -3 and the hypoteneuse is 5) and a triangle in Quadrant III (where y=-1 and x=-2 in terms of any diagram. We know y is in Quad III because it is >180 but <270.

We can calculate the hypoteneuse (using pythag) in Quad III as we have the other 2 sides ( because `tan y=1/2)`

`r^2 = x^2 + y^2`   and   `therefore r^2 = (-2)^2 + (-1)^2`

`therefore r=sqrt5`

To solve, also calculate the y value in Quad II. We know by calculation using Pythag  `y^2= 5^2-(-3)^2`

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`therefore y=sqrt16=4` and so  `sin x = 4/5 and`  `cos x=-3/5`

and we know `sin y = -1/sqrt5`   and that `cos y=-2/sqrt5`

`cos(x+y) = cos x.cos y - sinx.siny`

`therefore =( -3)/5 times(-2)/sqrt5 - 4/5 times (-1)/sqrt5`

`therefore = 6/(5sqrt5) - (-4)/(5sqrt5)`

`therefore = 10/(5sqrt5)`

Now simplify = `2/sqrt5`  and rationalize the denominator:

`therefore cos(x+y)= 2/sqrt5 times sqrt5/sqrt5`

`therefore = (2sqrt5)/5`

Ans:cos(x+y)`= (2sqrt5)/5`

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