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March 30, 2015

March 30, 2015

Posted by **katie** on Tuesday, November 29, 2011 at 5:12pm.

y=7cos(x-pi/6) for x=pi/6, pi/3, 2pi/3, pi, 7pi/6

- trig -
**Reiny**, Tuesday, November 29, 2011 at 5:17pmI will do the first one, you do the others.

Let me know what you get.

y = 7cos(x-π/6) , for x = π/6

= 7cos(π/6 - π/6)

= 7cos 0

= 7(1) = 7

one more... if x = 2π/3

y = 7cos(2π/3 - π/6)

= 7cos (π/2)

= 7(0) = 0

- trig -
**katie**, Tuesday, November 29, 2011 at 5:28pmi didn't realize it was just plugging in! thanks! i just need to figure out how to subtract fractions with pi in them now

- trig -
**katie**, Tuesday, November 29, 2011 at 5:37pmpi/3 is 7 square root 3/ 2 pi is - 7 square root 3 / 2 and 7pi/6 is -7

- trig -
**Reiny**, Tuesday, November 29, 2011 at 5:46pmI don't like the way you wrote those answers

Be mathematically correct, ..... e.g.

if x = π/3

y = 7cos(π/3 - π/6)

= 7 cos (π/6)

= 7(√3/2) = 7√3/2

if x = 7π/6

y = 7cos(7π/6 - π/6)

= 7cos π

= 7(-1) = -7

A statement such as "π/3 is 7√3/2" makes no sense

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