Evaluating cosine and sine HELP?

Question

Evaluate: 
cos(a − b) if cos a = 4/5 tan a < 0, tan b =−√15 and cos b < 0 

cos(a-b) = 

and 

Evaluate: 
sin(x + y) if sin x = 3/5 sec x > 0, cos y = −(2√5)/5 and tan y < 0. 

sin(x + y) = 
 

Answer 1

The value of  cos a = 4 / 5  and tan b = - √ 15

cos (a - b) = cos a cos b + sina sin b

cos a = adjacent side / hypotenuse    = 4 / 5

sin a = opposite side / hypotenuse

Hypotenuse ² = adjacent side ² + opposite side ²

5² = 4² + opposite side ²

opposite side  = 3.

Hence sin a = - 3 / 5.

tan b =- √ 15

tan b = opposite side  / adjacent side 

Hypotenuse ² = adjacent side ² + opposite side ²

Hypotenuse ² = (- √ 15) ² + 1 ²

Hypotenuse = 4.

sin b = √ 15 / 4 

cos b = - 1 / 4

cos (a - b) = cos a cos b + sina sin b

cos (a - b) =  (4 / 5) (- 1 / 4) + (-3 / 5) (√ 15 / 4 ) = ( - 1 / 5)  - ( 3√ 15 / 20 )

                  = (-4 - 3√ 15) / 20  .

Hence the value of cos (a - b) = (-4 -3√ 15) / 20  .

Answer 2

The value of  sin (x) = 3 / 5 

Now evaluating for cos x:

sin x = opposite side / hypotenuse

Hypotenuse ² = adjacent side ² + opposite side ²

 5² = adjacent side ² +3²   = > 25 - 9 =adjacent side ² =>adjacent side = 4.

cos x = 4 /5.

Since secx >0 , that is secx is possitive that is x lies in first quadrant .

The value of cos y = −(2√5)/5

Now evaluating for sin y:

cos y = adjacent side / hypotenuse =cos a =  −(2√5)/5 . 

Hypotenuse ² = adjacent side ² + opposite side ²

5²  = (−(2√5)) ²  + opposite side  ²

25 - (4/5) = opposite side  ²

opposite side = 11 /√5

Sice tan y< 0 tha t is tan y is negative ,y lies in fourth quadrant.

sin y = opposite side / hypotenuse (11/√5) / 5= 11  / (5√5) = 11√5 / 25

sin (x +y)  = sin x cos y + cos x sin y

sin (x + y) = (3 / 5)(−(2√5)/5) + (4 / 5) (11√5 / 25)

                 =( -6√5)/ 25   + (44 √5) / 125

             = (5 (-6√5) + 44 √5) / 125

            = (- 30 √5 + 44√5) / 125  

           = 14 √5 /125.

Hence the value of  sin (x +y) = 14 √5 /125.