SOLUTION: Find the smallest number Θ larger than 8pi such that sinΘ = (square root2/2)

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Question 1168755: Find the smallest number Θ larger than 8pi such that sinΘ = (square root2/2)
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Use a unit circle or reference table to see that
sin(pi/4) = sqrt(2)/2
note: pi/4 radians = 45 degrees

From here, add on 2pi to get a coterminal angle
pi/4 + 2pi = pi/4 + 8pi/4 = 9pi/4

Use your calculator to get
8pi = 25.1327412287183
9pi/4 = 7.06858347057703
So 9pi/4 is not larger than 8pi

We need to add on another multiple of 2pi
9pi/4 + 2pi = 9pi/4 + 8pi/4 = 17pi/4
17pi/4 = 13.3517687777567
which is not bigger than 8pi = 25.1327

So we need to add on yet another multiple of 2pi
We keep going until the result is larger than 8pi = 25.1327
17pi/4+2pi = 17pi/4+8pi/4 = 25pi/4
25pi/4 = 19.6349540849362
Still not big enough

Add on another 2pi
25pi/4+2pi = 25pi/4+8pi/4 = 33pi/4
33pi/4 = 25.918139
Now this is finally bigger than 8pi = 25.1327

Answer: theta = 33pi/4

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


sin%28x%29=sqrt%282%29%2F2 means a reference angle of 45 degrees, or pi/4. That means an angle of 2k(pi)+pi/4 or 2k(pi)+3pi/4.

8pi is an integer multiple of 2pi; the smallest angle greater than 8pi with a sine of sqrt(2)/2 is

8pi%2Bpi%2F4+=+33pi%2F4