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anonymous

  • 5 years ago

Use the fact that the trig functions are periodic to find the exact value of the given expression. sin ((17pi)/4) = ((sqrt{2})/2)

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  1. anonymous
    • 5 years ago
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    Is this answer correct?

  2. Mertsj
    • 5 years ago
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    yes

  3. Mertsj
    • 5 years ago
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    17pi/4 is the same as pi/4

  4. anonymous
    • 5 years ago
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    To clarify, the value of sin(17pi/4) is the same as sin(pi/4) ;) pi/4 radians is 45 degrees. Take a triangle ABC with A at the origin, B at (1,0), and C at (1,1). You'll have a right angle at B and 45 degree angles at A and C. AC will have a 45 degree angle over the x-axis. sin is the same as the ratio BC/AC. You know AB and BC are both 1 units long. Pythagoras knew that AB^2+BC^2 = AC^2, which in our case is 1^2+1^2 = AC^2, so AC = sqrt(2); since we know BC/AC is the same as sin(pi/4) (because AC's angle over the x axis is pi/4), it follows that BC/AC = sin(pi/4), or sin(pi/4) = 1/sqrt(2), which is sqrt(2)/2 :)

  5. myininaya
    • 5 years ago
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    17pi/4 is not the same as pi/4 but sin(17pi/4) is the same as sin(pi/4) since \[\sin(\frac{16}{4} \pi +\frac{1 }{4 }\pi)=\sin(2\pi+\frac{\pi}{4})=\sin(\frac{\pi}{4}) \text{ since twopi is just one rotation }\]

  6. anonymous
    • 5 years ago
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    For what numbers theta is f(theta)=tan(theta) not defined? f(theta)=tan(theta) is not defined for numbers that are (even multiples, odd multiples, multiples) of (45*, 90*, 180*). Select an answer from each set of parentheses.

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