Instructions: Find the exact value of the expression giventhat sec(x)= 3/2 and csc(y)=3 where x and y are in the 1st quadrant.
From what I have been able to deduce, cos(x)= 2/3 andsin(y)= 1/3. Because of this, I solved the following problems based on thisinformation. If anyone can double check and explain to me if I got somethingwrong, I would really appreciate it! I’m very bad at trig so I’m drowning inprecal right now in college.
From what I have been able to deduce, cos(x)= 2/3 andsin(y)= 1/3. Because of this, I solved the following problems based on thisinformation. If anyone can double check and explain to me if I got somethingwrong, I would really appreciate it! I’m very bad at trig so I’m drowning inprecal right now in college.
- Sin(x+y)
- Sin((2/3)+(1/3)) = sin(2/3)cos(1/3) +sin(1/3)cos(2/3)
- = (2/3)(1/3) + (1/3)(2/3)
- = (4/9)
- Sin((2/3)+(1/3)) = sin(2/3)cos(1/3) +sin(1/3)cos(2/3)
- Tan(x-y)
- tan((2/3)-(1/3)) = (tan(2/3) – tan(1/3))/(1 +tan(2/3)tan(1/3))
- = ((2/3) – (1/3))/(1 + (2/3)(1/3))
- = (1/3)/(11/9)
- =(3/11)
- tan((2/3)-(1/3)) = (tan(2/3) – tan(1/3))/(1 +tan(2/3)tan(1/3))
- Cos(y/2)
- Cos((1/3)/2)
-
- Cos((1/3)/2)
- Sin(2x)
- Sin(2(2/3))
- 2sin(2/3)cos(2/3)
- 2(2/3)(2/3)
- (8/9)
- Sin(2(2/3))