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Evalute in simplist radical form. Show all work including converting to degrees,...

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connorcolin3 | Valedictorian

Posted May 11, 2013 at 3:51 PM via web

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Evalute in simplist radical form. Show all work including converting to degrees, finding the reference angle and quadrant, etc.

(Cot7Pie/6)(Sec 4pie/3) 

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Tagged with algebra 2 trig, math

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 11, 2013 at 5:03 PM (Answer #1)

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You need to use the following trigonometric identities, such that:

`cot alpha = cos alpha/sin alpha`

`sec alpha = 1/cos alpha`

Reasoning by analogy, yields:

`cot((7pi)/6) = (cos((7pi)/6))/(sin((7pi)/6))`

`sec((4pi)/3) = 1/(cos((4pi)/3))`

You should write` (7pi)/6` as the summation `(2pi)/3 + pi/2` such that:

`cos((7pi)/6) = cos((2pi)/3 + pi/2)`

You need to expand `cos((2pi)/3 + pi/2)` using the formula, such that:

`cos(a + b) = cos a*cos b - sin a*sin b`

Reasoning by analogy yields:

`cos((2pi)/3 + pi/2) = cos((2pi)/3)*cos (pi/2) - sin((2pi)/3)*sin (pi/2)`

Since `cos pi/2 = 0` and `sin pi/2 = 1` yields:

`cos((2pi)/3 + pi/2) = -sin((2pi)/3)`

You need to expand `sin ((2pi)/3 + pi/2)` using the formula, such that:

`sin(a + b) = sin a*cos b + sin b*cos a`

Reasoning by analogy yields:

`sin ((2pi)/3 + pi/2)= sin((2pi)/3)*cos pi/2 + sin pi/2*cos ((2pi)/3)`

`sin ((2pi)/3 + pi/2)= cos ((2pi)/3)`

Replacing -`sin((2pi)/3)` for `cos((7pi)/6)` and `cos ((2pi)/3)` for `sin((7pi)/6) ` yields:

`(cot((7pi)/6))/(sec((4pi)/3)) = ((-sin((2pi)/3))/(cos ((2pi)/3)))/(1/sin((4pi)/3))`

You need to use the double angle identity for `sin((4pi)/3)` , such that:

`sin((4pi)/3) = 2 sin((2pi)/3)*cos((2pi)/3)`

Replacing `2 sin((2pi)/3)*cos((2pi)/3)` for `sin((4pi)/3)` yields:

`(cot((7pi)/6))/(sec((4pi)/3)) = ((-sin((2pi)/3))/(cos ((2pi)/3)))/(1/(2 sin((2pi)/3)*cos((2pi)/3)))`

Reducing duplicate terms yields:

`(cot((7pi)/6))/(sec((4pi)/3)) = -2sin^2((2pi)/3)`

You need to use the double angle formula to evaluate s`in ((2pi)/3)` ,such that:

`sin ((2pi)/3) = 2 sin(pi/3)*cos(pi/3)`

`sin ((2pi)/3) = sin 120^o = 2 sin 60^o*cos 60^o`

`sin ((2pi)/3) = 2*sqrt3/2*1/2 => sin^2 ((2pi)/3) = 4*3/4*1/4 = 3/4`

`(cot((7pi)/6))/(sec((4pi)/3)) = -2*(3/4) => (cot((7pi)/6))/(sec((4pi)/3)) = -3/2`

Hence, evaluating `(cot((7pi)/6))/(sec((4pi)/3))` yields `-2sin^2((2pi)/3) = -3/2` , where `(2pi)/3 = 120^o` is an angle in quadrant 2, where the sine function values are positive.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 11, 2013 at 5:48 PM (Answer #2)

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Evaluate `(cot(7pi)/6)(sec(4pi)/3)` :

(1)`cot(7pi)/6=(cos((7pi)/6))/(sin((7pi)/6))`

`(7pi)/6` is in the third quadrant with reference angle `pi/6` ; in the third quadrant both sine and cosine are negative. `sin(pi/6)=1/2,cos(pi/6)=sqrt(3)/2`

Then `cot(7pi)/6=(-sqrt(3)/2)/(-1/2)=sqrt(3)`

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(2)`sec(4pi)/3=1/(cos((4pi)/3))`

`(4pi)/3` is in the third quadrant with a reference angle of `pi/3` ; in the third quadrant cosine is negative. `cos(pi/3)=1/2`

Then `sec((4pi)/3)=1/(-1/2)=-2`

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Thus `(cot(7pi)/6)(sec(4pi)/3)=sqrt(3)*(-2)=-2sqrt(3)`

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