Asked by Brendan
Find sin(x+y), cos(x-y), tan(x+y), and the quadrant of (x+y) if sinx= -1/4, cosy= -4/5, with x and y in quadrant 3.
Answers
Answered by
Reiny
you need to sketch right-angled triangles to get the missing sides
given: sinx = -1/4, y = -1, r = 4
x^2 + y^2 = r^2
x^2 + 1 = 16
x = -√15 in III
So sinx = -1/4 , cosx = -√15/4 , tanx = 1/√15
given: cosy = -4/5, x = -4, r = 5
then y = -3
siny = -3/5, cosy = -4/5 , tany = 3/4
Now we have all the parts, let's just use the definitions of ....
sin(x+y) = sinxcosy + cosxsiny
= (-1/4)(-4/5)+ (-√15/4)(-3/5)
= (4 + 3√15)/20
cos(x-y) = cosxcosy + sinxsinx
= .....
you try this one
tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (1/√15 + 3/4)/(1 - (1/√15)(3/4))
= ( (4+3√15)/(4√15) )/( (4√15 - 3)/(4√15))
= (4 + 3√15)/(4√15 - 3)
given: sinx = -1/4, y = -1, r = 4
x^2 + y^2 = r^2
x^2 + 1 = 16
x = -√15 in III
So sinx = -1/4 , cosx = -√15/4 , tanx = 1/√15
given: cosy = -4/5, x = -4, r = 5
then y = -3
siny = -3/5, cosy = -4/5 , tany = 3/4
Now we have all the parts, let's just use the definitions of ....
sin(x+y) = sinxcosy + cosxsiny
= (-1/4)(-4/5)+ (-√15/4)(-3/5)
= (4 + 3√15)/20
cos(x-y) = cosxcosy + sinxsinx
= .....
you try this one
tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (1/√15 + 3/4)/(1 - (1/√15)(3/4))
= ( (4+3√15)/(4√15) )/( (4√15 - 3)/(4√15))
= (4 + 3√15)/(4√15 - 3)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.