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April 18, 2014

April 18, 2014

**Recent Homework Questions About Trigonometry**

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**trig**

cos4x = 2cos^2(2x)-1 = 2(2cos^2(x)-1)^2 - 1 just expand that sin(x/4) = sqrt((1-cos(x/2))/2) = sqrt((1-sqrt((1-cos(x))/2)/2
*Tuesday, May 29, 2012 at 3:56am*

**trig**

1) Use double-angle identities to write the following expression, using trigonometric functions of x instead of 4x. cos 4x 2) Use half-angle identities to write the following expression, using trigonometric functions of x instead of x/4. sin x/4
*Monday, May 28, 2012 at 11:36pm*

**trig**

We form a rt. triangle: X = 6 meters = Hor. Y = Ver. = Dist. from gnd to junction of wire and pole. Z = Hyp. = Length of the wire. A = Angle bet. gnd and wire. B = Angle bet. pole and wire. A + B = 90 Deg. A = 3B. 3B + B = 90 B = 22.5 Deg. A = 3B = 3*22.5 = 67.5 Deg. Z*cosA = ...
*Monday, May 28, 2012 at 8:03pm*

**trig**

4 divided by 6 = 1.5 1.5 times 18 = 27 the answer is : 27
*Sunday, May 27, 2012 at 11:20pm*

**trig**

AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50° (a) the length BC (b) The size of angle ABC; (c) The size of angle CAD (d) The area of triangle ACD
*Sunday, May 27, 2012 at 5:08pm*

**trig**

An electric pole is supported to stand vertically on a level ground by a tight wire. The wire is pegged at a distance of 6 meters from the foot of the pole. The angle that the wire makes the ground is three times the angle it makes with the pole.
*Sunday, May 27, 2012 at 4:35pm*

**trig**

plot of land ABCD such that AB = 85m, BC = 75m, CD = 60m, DA = 50m and angle ACB = . Determine the area of the plot in hectares correct to two decimal places
*Sunday, May 27, 2012 at 4:33pm*

**algebra 2/trig**

x+5 1 ------- - 1= ------- x^2-2x x^2-2x
*Sunday, May 27, 2012 at 1:23pm*

**Trig **

i dot geht thhis werlk? cahn sumwone elp meh?
*Saturday, May 26, 2012 at 2:52pm*

**Trig Help Please!!!**

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
*Saturday, May 26, 2012 at 2:50pm*

**Calc- Trig Substitution**

hmm. upon rereading your answer, you are in fact correct.!!
*Friday, May 25, 2012 at 5:19am*

**Calc- Trig Substitution**

not even close :-( let x = 8secu dx = 8secu*tanu du x^2-64 = 64sec^2 u - 64 = 64tan^2 u dx/√(x^2-64) = 8*secu*tanu/8tanu = secu since ∫secu = ln|secu+tanu|, we have Recall that x = 8secu, so secu = x/8 tanu = √(x^2/64 - 1) = 1/8 √(x^2-64) ∫secu = ...
*Friday, May 25, 2012 at 5:18am*

**Calc- Trig Substitution **

integral of dx/square root(x^2-64) My answer is: ln (x/8 + square root(x^2-64)/8) +c is this right?
*Friday, May 25, 2012 at 2:23am*

**trig**

Using sin^2 x = 1-cos^2(x) the equation becomes: cos²(x)+cos(x)-2=0 Use the substitution c=cos(x) to transform the equation to : c²+c-2=0 (c+2)(c-1)=0 c=-2 or c=+1 Since cos(x) cannot equal -2, solution is rejected. Now solve for all values of 0≤x≤2π where ...
*Thursday, May 24, 2012 at 8:07am*

**trig-math**

A stairway must be built to a deck that is 20 feet above ground level. To the nearest half foot, how far from the base of the deck, on ground level, should the beginning of the stairway be placed so that the stairway forms a 60° angle from the ground?
*Thursday, May 24, 2012 at 6:13am*

**Trig**

yes. sorry. I forgot the comma
*Wednesday, May 23, 2012 at 11:37pm*

**trig**

Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=pi, x=pi/2, x= 2pi/3 b. x=3pi/7, x=pi/2, x=2pi/3 c. x=3pi/7, x=3pi/2, x=3pi/2 d. x=pi, x=pi/2, x=3pi/2
*Wednesday, May 23, 2012 at 11:12pm*

**trig**

let x = height of tree tan(angle) = x/27 tan(26) = x/27 x = 0.488*27 x = 13.17 = 13 m
*Wednesday, May 23, 2012 at 10:56pm*

**trig :)**

You're welcome!
*Wednesday, May 23, 2012 at 10:29pm*

**trig**

thank you so much, it helps! :)
*Wednesday, May 23, 2012 at 10:17pm*

**trig**

tan(x) has a period of π, so tan(x)=tan(x+π), in fact, tan(x+kπ)=tan(x), where k∈Z So to evaluate tan(257π/4) =tan((257/4)π) =tan(64π + π/4) =tan(π/4) I will let you figure out which answer to choose.
*Wednesday, May 23, 2012 at 10:11pm*

**trig**

evaluate the expression tan (257pi/4) a.-1 b. -(sqrt2)/2 c. 1 d. (sqrt2)/2
*Wednesday, May 23, 2012 at 9:44pm*

**trig**

A tree casts a shadow of 27 meters when the angle of elevation of the sun is 26 degrees. Find the height of the tree to the nearest meter. a. 24 m b. 15 m c. 320 m d. 13 m
*Wednesday, May 23, 2012 at 9:40pm*

**trig**

s = rθ 1.1 = 3θ θ = .366666 radians = 21°
*Wednesday, May 23, 2012 at 11:52am*

**trig**

a voltmeter's pointer is 3 cm in length. find the degrees through which is rotates when it moves 1.1 cm on the scale?
*Wednesday, May 23, 2012 at 11:11am*

**Trig**

look at the solution I gave here to a similar problem http://www.jiskha.com/display.cgi?id=1267493976
*Wednesday, May 23, 2012 at 5:01am*

**Trig**

??? Is that a polar coordinate point? is there a missing comma? did you mean (-2 , π/6) ?
*Wednesday, May 23, 2012 at 4:58am*

**trig**

cos 2x + √2/2 =0 cos 2x = -√2/2 I know cos π/4 = √2/2 or cos 45° = +√2/2 but 2x could be in II or III 2x = π-π/4 or 2x = π + π/4 2x = 3π/4 or 2x = 5π/4 x = 3π/8 or x = 5π/8 ---> 67.5° or 112.5°...
*Wednesday, May 23, 2012 at 4:56am*

**trig**

b) to multiply two complex numbers in complex form if u = r1(cos Ø1 + isinØ1) and v = r2(cosØ2 + isinØ2) then uv = r1r2(cos(Ø1+Ø2) + isin(Ø1+Ø2) and u/v = r1/r2 (cos(Ø1-Ø2) + isin(Ø1-Ø2) so (z...
*Wednesday, May 23, 2012 at 4:31am*

**trig**

7. z = -5√3/2 + 5/2i = 5(-√3/2 + (1/2)i ) argument : tan Ø = (1/2) / (-√3/2), where Ø is in II tan Ø = -1/√3 Ø = 150° or 5π/6 radians z = 5(cos 150° + isin 150°) or 5(cos 5π/6 + isin 5π/6) in the ...
*Wednesday, May 23, 2012 at 4:23am*

**trig**

6. I will do E, you try the others the same way let z = -5 + 5i modulus z = |z| = √((-5)^2 + 5^2) = √50 = 5√2 argument: tanØ = 5/-5 = -1, where Ø is in quad II Ø = 135° G. is done the same way for D, think of it as -5 + 0i
*Wednesday, May 23, 2012 at 4:11am*

**trig**

6. Compute the modulus and argument of each complex number. I did a-c and f. D. -5 E. -5+5i G. -3-4i 7. Let z= -5sqrt3/2+5/2i and w= 1+sqrt3i a. convert z and w to polar form b. calculate zw using De Moivres Theorem c. calculate (z/w) using De M's theorem Please help with ...
*Wednesday, May 23, 2012 at 12:41am*

**trig**

Find all solutions of cos 2x + √2/2 =0. a) + or - 3/8pi +Kpi b) 2/3pi + Kpi, 5/3pi + Kpi c) + or - 2/5 +Kpi d) 2/3 pi + 2Kpi, 5/3pi + 2Kpi I really need help on this question :(
*Tuesday, May 22, 2012 at 11:27pm*

**Trig**

what are the rectangular coordinates of (-2(pi/6))? please help.
*Tuesday, May 22, 2012 at 11:25pm*

**math trig**

ln(2x-5) = 4 we raise both sides by e: e^(ln(2x-5)) = e^4 the e and ln will be cancelled, leaving only the 2x - 5: 2x - 5 = e^4 2x = e^4 + 5 x = (e^4 + 5)/2 hope this helps~ :)
*Monday, May 21, 2012 at 10:55pm*

**math trig**

ln(2x-5)= 4 how do i this problem? thanks!
*Monday, May 21, 2012 at 10:45pm*

**TRIG**

I need to state the period and 2 consecutive asymptotes on the graph for the following questions. 1: y = -3 tan pi*x period: pi (?) asymptotes: ? 2: y = 2 sec 4x period: ? asymptotes: ? 3: y = csc (x/3) period: ? asymptotes: ? 4: y = 3 cot (pi*x/2) period: ? asymptotes: ?
*Monday, May 21, 2012 at 6:03pm*

**geometry**

did you make a sketch?? This is just right-angled triangle trig tan 29.78° = h/980 h = 980tan29.78° =appr 561 feet so when the height is 280.5 feet tan Ø = 280.5/980 =.28612.. Ø = 15.97°
*Monday, May 21, 2012 at 8:05am*

**graphing trig functions**

I see the equation of a straight line, not a trig function
*Monday, May 21, 2012 at 7:53am*

**trig**

4 cos theta - 1
*Sunday, May 20, 2012 at 11:31pm*

**graphing trig functions**

given y=(3x+2pi) how would i graph one period this without a calculator? amplitude:1 period: 2pi/b = 2pi/3 is this correct? what is the minimum and maximum points?
*Sunday, May 20, 2012 at 10:15pm*

**trig**

how can you find points of a tri equation at orgin 1/4, 1/2, 3/4, and end of the period
*Sunday, May 20, 2012 at 9:53pm*

**Trig**

sin θ= sqrrt3/2 This is just your standard pi/3. So, pi/3 and 2pi/3 b. 4pi/3, 5pi/3
*Saturday, May 19, 2012 at 11:10pm*

**Trig**

Find two values of theta that satisfy the equation. give your answers in degrees (0degrees <theta<360degrees ) and radians (0<theta<2pi) a. sin θ= sqrrt3/2 b.sin θ= - sqrrt3/2
*Saturday, May 19, 2012 at 10:44pm*

**Trig**

s = rθ θ = s/r = 10/16 = 5/8
*Friday, May 18, 2012 at 2:09pm*

**Trig**

Find the radian measure of the central angle of a circle of radius r that intercepts an arc of length s.? Radius, r=16 ft Arc length, s=10 ft
*Friday, May 18, 2012 at 12:51pm*

**graphing trig functions**

In that case, pick points that you know. For 2sinx, say, you know that sin 0 = 0 sin pi/6 = .5 sin pi/4 = .707 sin pi/3 = .866 sin pi/2 = 1 and so on If you know a few key values, you can sketch the graph easily. If you don't happen to know the value of pi/3, etc, use ...
*Friday, May 18, 2012 at 10:18am*

**graphing trig functions**

but for my quiz, i am not allowed a calculator
*Friday, May 18, 2012 at 12:58am*

**graphing trig functions**

You can pick as many x points as you want. The more you plot, the easier it is to plot a smopoth and accurate curve. For the y values that go with each x, you need a calculator, a special slide rule or a table of trig functions. Hardly anyone uses tables or slide rules anymore.
*Friday, May 18, 2012 at 12:45am*

**graphing trig functions**

for graphing basic trig functions such as y=2sinx or y=1/3cosx, how do you know what the points are to graph with out using a calculator?
*Friday, May 18, 2012 at 12:31am*

**trig**

sinAcosB = 1/2 (sin(A+B) + sin(A-B)) Looks like 1/2(sin6x+sin2x)
*Thursday, May 17, 2012 at 11:46pm*

**trig**

write sin4xcos2 as the sum or difference of two functions. answers: 1/2(cos6x+cos2x), 1/2(cos2x-cos6x), 1/2(sin6x+sin2x), sin6x-sin2x
*Thursday, May 17, 2012 at 9:38pm*

**trig**

U∘V = |U| |V|cos Ø, where Ø is the angle between 2 - 15 = √13√26cosØ cosØ = -13/13√2 = -1/√2 Ø = 135°
*Thursday, May 17, 2012 at 8:07pm*

**trig function**

Oh your right! Now I see that, thanks so much!
*Thursday, May 17, 2012 at 7:56pm*

**trig**

The coefficients of sin(x) and -cos(x) are 1 and -1 respectively. We divide each term by the factor sqrt(1²+(-1)²)=sqrt(2) y=sinx-cosx => y=sqrt(2)[sin(x)*(1/sqrt(2)+cos(x)*(-1/sqrt(2)) Since cos(-π/4)=1/sqrt(2), and sin(-π/4)=-1/sqrt(2), we set a=3&...
*Thursday, May 17, 2012 at 7:46pm*

**trig function**

(fxg)(x) = (cosx - sinx)(cosx + sinx) = cos^2 x - sin^2 x = cos 2x , by definition. notice the period of the new one is 2π/2 = π while the period of the first one was 2π notice the compression by a factor of 1/2 ?
*Thursday, May 17, 2012 at 7:46pm*

**trig**

write sin4xcos2 as the sum or difference of two functions. answers: 1/2(cos6x+cos2x), 1/2(cos2x-cos6x), 1/2(sin6x+sin2x), sin6x-sin2x
*Thursday, May 17, 2012 at 7:18pm*

**trig**

find the angle between vector U=<2,3> and V=<1,-5> answers: 88 degrees, 45 degrees, 135 degrees, or 92 degrees...?
*Thursday, May 17, 2012 at 7:13pm*

**trig function**

Suppose f(x)=cos x - sin x and g(x)=cos x + sin x. Explain why the graph of (fxg)(x) is equivalent to the graph of h(x)=cos x after it has been horizontally compressed by a factor of 1/2. Please help, thank you!
*Thursday, May 17, 2012 at 7:10pm*

**trig**

ksin(x+a) = k(sinxcosa) + cosxsina) = ksinxcosa + kcosxsina then ksinxcosa + kcosxsina = sinx - cosx , which must be an identity, so true for all x let x =0 k(0)cosa + k(1)sina = 0-1 ksina= 1 sina = 1/k let x = 90° k(1)(cosa) + k(0) = 1-0 cosa = 1/k sina/cosa = (1/k) / (1/...
*Thursday, May 17, 2012 at 7:03pm*

**trig**

write y=sinx-cosx in the form y=ksin (x+a), where the measure of a is in radians. answers: sqrt2sin (x+3pi/4), sqrt2sin (x+5pi/4), sqrt2sin (x+pi/4), or sqrt2sin (x+7pi/4) ...?
*Thursday, May 17, 2012 at 6:49pm*

**trig**

write sin4xcos2 as the sum or difference of two functions. answers: 1/2(cos6x+cos2x), 1/2(cos2x-cos6x), 1/2(sin6x+sin2x), sin6x-sin2x
*Thursday, May 17, 2012 at 6:42pm*

**trig**

when t=0 , all answers with sin are zero all answers with cosine are 4, so we eliminate the sine answers period of a sine or cosine function is 2π/k but we are told: 2π/k = 2 k = π so it must be y = 4cos πt
*Thursday, May 17, 2012 at 6:30pm*

**trig**

9π/5 = 2π + π/5 so the "angle in standard position" or the "reference angle"is π/5
*Thursday, May 17, 2012 at 6:27pm*

**trig**

make a sketch of a 2-3-√13 right-angled triangle to see that sin[tan^-1 (2/3)] = 2/√13
*Thursday, May 17, 2012 at 6:25pm*

**trig**

Write an equation for simple harmonic motion, given that the amplitude equals 4 inches and the period equals 2 seconds and the max displacement occurs when t=0. answers: y=4sin pi t, y=4cos pi t, y=4cos t, y=4sin2 pi t
*Thursday, May 17, 2012 at 6:24pm*

**trig**

an angle with the radian measure of 9pi/5. find radian measure of its reference angle. answers: -4/5, 4pi/5, -pi/5, or pi/5
*Thursday, May 17, 2012 at 6:21pm*

**trig**

recall cosAcosB - sinAsinB = cos(A+B) then cos6xcosx - sin6xsinx = cos(6x+x) = cos(7x)
*Thursday, May 17, 2012 at 6:20pm*

**trig**

what is the exact value of sin[tan^-1(2/3)]
*Thursday, May 17, 2012 at 6:19pm*

**trig**

sin a = 15/17 , with a in I, so cos a = 8/17 cos b = 4/5, with b in IV, then sin b = -3/5 cos(a+b) = cosa cosb - sina sinb = (8/17)(4/5) - (15/17)(-3/5) = (32 + 45)/85 = 77/85
*Thursday, May 17, 2012 at 6:19pm*

**trig**

would that be 88 degrees, 45 degrees, 135 degrees, or 92 degrees?
*Thursday, May 17, 2012 at 6:10pm*

**trig**

cos T = U dot V / ( |U|* |V| ) = (2-15) / (sqrt 13 + sqrt 26) = -13 / (sqrt 13 + sqrt 2 sqrt 13) = -13/ [sqrt 13 (1 + sqrt 2) ]
*Thursday, May 17, 2012 at 6:04pm*

**trig**

Yes
*Thursday, May 17, 2012 at 6:00pm*

**trig**

sqrt (4 + 9) = sqrt 13 tan T = -2/-3 = 33.7 deg in quad 3 so 33.7 + 180 = 213.7 degrees
*Thursday, May 17, 2012 at 6:00pm*

**trig**

16 cos 120 i + 16 sin 120 j cos 120 = -.5 sin 120 = +.866 or (1/2)sqrt 3 so -8 i + 8 sqrt 3 j SKETCH it on a GRAPH !!!!
*Thursday, May 17, 2012 at 5:57pm*

**trig**

would the 3 answers be: pi/6, 5pi/6, 3pi/2
*Thursday, May 17, 2012 at 5:43pm*

**trig**

2 U = 10 i + 22 j 7 V = 14 i - 21 j ----------------- subtract for 2 U - 7 V = -4 i + 1 j but I have no idea if that is what you want. Please repair typo.
*Thursday, May 17, 2012 at 5:41pm*

**trig**

its 2U-7V
*Thursday, May 17, 2012 at 5:39pm*

**trig**

2U-7U must be a TYPO maybe 2 U - 7V ? or 2V - 7 U ???
*Thursday, May 17, 2012 at 5:37pm*

**trig**

thats the problem i have..? the answers may be: 24i+43, -4i+j, -4i+43j, or 24i+j
*Thursday, May 17, 2012 at 5:36pm*

**trig**

Corection: sin ( x ) = 1 / 2 sin ( x ) = - 1
*Thursday, May 17, 2012 at 5:34pm*

**trig**

cos ^ 2 ( x ) = 1 - sin ^ 2 ( x ) 2 cos ^ 2 ( x ) - sin ( x ) = 1 2 * [ 1 - sin ^ 2 ( x ) ] - sin ( x ) = 1 2 - 2 sin ^ 2 ( x ) - sin ( x ) = 1 Subtract 1 to both sides 2 - 2 sin ^ 2 ( x ) - sin ( x ) - 1 = 1 - 1 - 2 sin ^ 2 ( x ) - sin ( x ) + 1 = 0 Substitute : sin ( x ) = u...
*Thursday, May 17, 2012 at 5:33pm*

**trig**

2 + 2 i sqrt 3 ???? 2 (1 + i sqrt 3) 4 (1/2 + i sqrt(3) /2 ) 4 (cos 60 deg + i sin 60 deg) 4 cis 60 or 4 cis (pi/3)
*Thursday, May 17, 2012 at 5:23pm*

**trig**

U=5i+11j and V=2i-3j, find 2U-7U
*Thursday, May 17, 2012 at 5:02pm*

**trig**

write 2+2 squareroot of 3i in trig form.
*Thursday, May 17, 2012 at 5:01pm*

**trig**

find the angle between vectors U= <2,3> and V= <1, -5>
*Thursday, May 17, 2012 at 4:53pm*

**trig**

a vector has a magnitude of 16 and direction of 120 degrees. write the vector in terms of unit vecotrs i and j.
*Thursday, May 17, 2012 at 4:52pm*

**trig**

what is the magnitude and direction of the vector U= -2i - 3j
*Thursday, May 17, 2012 at 4:50pm*

**trig**

write y=sinx-cosx in the form y=ksin (x+a), where the measure of a is in radians.
*Thursday, May 17, 2012 at 4:48pm*

**trig**

solve 2cos^2x-sinx=1, 0 < or equal to x < 2pie
*Thursday, May 17, 2012 at 4:47pm*

**trig**

write sin4xcos2 as the sum or difference of two functions.
*Thursday, May 17, 2012 at 4:45pm*

**trig**

what's the single trig. function of cos6xcosx-sin6xsinx
*Thursday, May 17, 2012 at 4:44pm*

**trig**

Given sin a = 15/17 with a in Quadrant I and cos beta = 4/5 with beta in QuandrantIV, find the exact value of cos(a+beta)
*Thursday, May 17, 2012 at 4:43pm*

**trig function**

Suppose f(x)=cos x - sin x and g(x)=cos x + sin x. Explain why the graph of (fxg)(x) is equivalent to the graph of h(x)=cos x after it has been horizontally compressed by a factor of 1/2. Thanks so much...
*Thursday, May 17, 2012 at 2:51pm*

**trig function**

Suppose f(x)=cos x - sin x and g(x)=cos x + sin x. Explain why the graph of (fxg)(x) is equivalent to the graph of h(x)=cos x after it has been horizontally compressed by a factor of 1/2. Thanks so much...
*Thursday, May 17, 2012 at 2:51pm*

**trig function**

Suppose f(x)=cos x - sin x and g(x)=cos x + sin x. Explain why the graph of (fxg)(x) is equivalent to the graph of h(x)=cos x after it has been horizontally compressed by a factor of 1/2. Thanks so much...
*Thursday, May 17, 2012 at 2:50pm*

**trig**

C = 180 - 81 - 50 = 49° b/sin81 = 12/sin49 b = 12sin81/sin49 = ... find a the same way
*Wednesday, May 16, 2012 at 11:36pm*

**trig**

let y = -3sin(2t - π/3) = -3sin 2(t - π/6) amplitude = |-3| = 3 period = 2π/2 = π phase shift: π/6 to the right
*Wednesday, May 16, 2012 at 11:33pm*

**trig**

what is the amplitude, phase shift, and period of -3sin(2t- pie/3)
*Wednesday, May 16, 2012 at 10:56pm*

**trig**

on a triangle, A = 50 degrees, B = 81 degrees, and c = 12. what does a, b, and C equal?
*Wednesday, May 16, 2012 at 10:54pm*

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