SOLUTION: Please help me solve Find the exact value of a. sin 300° b. tan(405°)(Hint:405° =360° +45°) c. cot (13π/3) (Hint:13π/3= 4π+π/3)

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve Find the exact value of a. sin 300° b. tan(405°)(Hint:405° =360° +45°) c. cot (13π/3) (Hint:13π/3= 4π+π/3)       Log On


   



Question 980061: Please help me solve
Find the exact value of
a. sin 300°
b. tan(405°)(Hint:405° =360° +45°)
c. cot (13π/3) (Hint:13π/3= 4π+π/3)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two parts to get you started.

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a)

300 degrees is in quadrant IV. The reference angle formula for quadrant IV is

R = 360 - x

where R is the reference angle and x is the given angle, so,

R = 360 - x
R = 360 - 300
R = 60

Now use either a 30-60-90 triangle, or a unit circle to find that sin%2860%29+=+%28sqrt%283%29%29%2F2

Because this angle (300 degrees) is in Q4, we know sine is negative

sin%28300%29+=+-sin%2860%29+=+-%28sqrt%283%29%29%2F2

Final Answer: sin%28300%29+=+-%28sqrt%283%29%29%2F2
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b)

tan%28a%2Bb%29+=+%28tan%28a%29%2Btan%28b%29%29%2F%281-tan%28a%29%2Atan%28b%29%29

tan%28360%2B45%29+=+%28tan%28360%29%2Btan%2845%29%29%2F%281-tan%28360%29%2Atan%2845%29%29

tan%28360%2B45%29+=+%280%2B1%29%2F%281-0%2A1%29

tan%28360%2B45%29+=+%280%2B1%29%2F%281-0%29

tan%28360%2B45%29+=+1%2F1

tan%28360%2B45%29+=+1

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So, tan%28360%2B45%29+=+tan%28405%29+=+1