If sinA=3/5 when pi/2 < A < pi and cosB=5/13 when 3pi/2 < B < 2pi, find the exact value of the function cos(5pi/6+B)
1,475 results
Calculus
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Algebra 2
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Algebra 2
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Math
If A and B are acute angle such that SinA=8/17 and CosB=3/5.Find 1, Cos(A+B) 2, Sin(A+B) 3, Sin(AB)

Math
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PreCalculus
If sinA=3/5 when pi/2 < A < pi and cosB=5/13 when 3pi/2 < B < 2pi, find the exact value of the function cos(5pi/6+B).

Trigonometry
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Precalculus
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APCalculus
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trig
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Trigonometry
If sinA=3/5 and cosB=5/13 and if A and B are measures of two angles in Quadrant I, find the exact value of the following functions. cotB = sin2A= 3)cos(5pi/6 + B) = tan(A  pi/4) =

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PreCalculus
Hello, I need a bit of help with the following problem: If sinA=3/5 when pi/2 < A < pi and cosB=5/13 when 3pi/2 < B < 2pi, find the exact value of the following equations: cotB = sin2A = tan(Api/4)=

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Find all solutions of the equation in the interval [0,2pi) 2 cos^2 xcos x = 0 2cos^2 + cosx + 0 (x+1/2) (x+0/2) (2x+1) (x+0) 1/2,0 2Pi/3, 4pi/3, pi/2, 3pi/2 my teacher circled pi/2 and 3pi/2 What did I do wrong? I don't understand...

precalculus
Given sina= 3/4 and cosB = (5)/13 and this is Quad II find cos(a+B)

trig
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4. Write the equation of the cosine function with an amplitude of 1/3, a period of 2pi, and a phase shift of 3pi/2. I got 1/3COS(1/2x3pi), but I really do not think that is correct D: any tips or advice?

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1. (P 15/17, 8/17) is found on the unit circle. Find sinΘ and cosΘ Work: P= (15/17, 8/17) cosΘ = a value P = (a,b) sinΘ = b value Answer: cosΘ = 15/17 sinΘ = 8/17 2. Should the triangle be solved beginning with Law of Sines or Law of Cosines.

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math
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (ab) = sina cosb  cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The solution to sin x =

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If the function is: f(x)=2cos(x)1 [0,2pi] A) What are the intervalos of concacity? B) What are the inflection pinta My answers: A) concave down: [0,pi/2) and (3pi/2, 2pi] concave up: (pi/2, 3pi/2) B) inflection points: (pi/2, 1) (3pi/2, 1) Please check

CALC HW
FIND THE X VALUES OF ALL POINTS, 2PI LESS THAN OR EQUAL TO X LESS THAN OR EQUAL TO 2PI, WHERE Y=X+B IS TANGENT TO Y=X+BSINX. I KNOW THE ANSWER IS PI/2 AND 3PI/2, BUT WHY AREN'T THEY PLUS OR MINUS. If they are tangent, they have the same coordinates, and

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Approximate the equation's solutions in the interval (0,2pi) sin2x sinx = cosx 2cos(x) (1/2sin^2x) = 0 Then I got 3pi/2, pi/2, pi/6 and 5pi/6 Then I substituted 03 and got 3pi/2 , 5pi/2 , 9pi/2 , pi/2, pi/6, 7pi/6, 13pi/6 , 19pi/6 , 5pi/6 , 11pi/6 ,

precal desperate
Given sina= 3/4 and cosB = (5)/13 and this is Quad II find cos(a+B)

maths
Please can you help me with this question? Choose the option which is a false statement: A arctan(tan2/3pi))=1/3pi B arccos(cos(3/4pi))=3/4pi C sin(arcsin(1/2pi))=1/2pi D arcsin(1/2squareroot3)=1/3pi E arcsin(sin(3/4pi))=1/4pi F arccos

Math
If sinA+sinB=1/2 and cosA+cosB=1/2 then the value of tanA+tanB

mathematics
if sinA+sinB=x and cosAcosB=y then find tan[(AB)/2]

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What values for theta(0

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trig
if tan x= 4/3 and pip < x < 3pi/2, and cot y= 5/12 with 3pi/2 < y < 2pi find sin(xy)

Math2
IF (1/SinA)+(1/cosA)= (1/sinB)+(1/cosB) Prove that cot((A+B)/2)=tanA tanB Please helppppppppppp

pre cal
cosA (7/25) pi < A < 3pi/2 sin B= 3/5 3pi/2

Precalculus
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given the graph of f(x) = x sinx, 0

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Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]? 4? ...what about 4cos2θ = 8cos^2 2θ? 2. True or False: sin^2 4x = 1 has 8 solutions on the interval [0, 2pi]?

Maths
Hi, can someone give me the answer to this question: secAtanA=1/4. Find cosA I got: CosA=4(1sinA) Is my answer good? Now multply each side by (1+sinA) CosA(1+sinA)=4 cos^2 A CosA=1/4 (1+sinA) 16(1sinA)=1+sinA 15=17SinA So you have sinA, now find Cos A

Math
If A and B are acute angle such that SinA=8/17 and CosB=3/5.Find 1, Cos(A+B) 2, Sin(A+B) 3, Sin(AB)

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Use the given information to evaluate cos (ab). Cot a = 3/4, cot b = 24/25; pi

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1) The period of a trig. function y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct? 2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after

PreCal(Please check)
1) Find the Asymptotes: y = 2 sec 2pi x pi/2b and 3pi/2b pi/2(2pi) = pi/4 and 3pi/4 Is this correct? 2) Find the period and asymptotes: y = 1/2 sec pi x / 2 b = pi/2 I do not know how to find the period of asymptotes for this one. My original answer was 1

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Algebra2
Find the values of the inverse function in radians. 1. sin^1(0.65)? answer choices: a. 0.71+2pi n and 0.71+2pi n b. 0.71+2pi n and 3.85+2pi n c. 0.86+2pi n and 0.86 +2pi n d. 0.61+2pi n and 2.54+2pi n My answer is B

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For questions 2 and 3, use a calculator to find the values of the inverse function in radians. sin^1(0.65) 0.71 + 2pi n and –0.71 + 2 pi n 0.71 + 2 pi n and –3.85 + 2 pi n 0.86 + 2pi n and –0.86 + 2pi n –0.61 + 2pi n and 2.54 + 2 pi n

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In the interval 0

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Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole. r=16cos 3Q r=16cos(3pi/3 + 3h) = 16(cos3pi/2 3h sin 3pi/2 sin h) cos 3pi/2 = 0 sin 3pi/2 = 1 Not symmetrical My teacher said that this was wrong!Why?

Trigonometric Identities
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i need someone to explain to me how i got this answer...i know the answer but i need an explanation to the problem... C=2pi(r) the answer is r=C/2pi...i need someone to explain it to me..thank you C = 2pi(r) Divide both sides by 2pi C/2pi = r > r =

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Trig
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Triangle HAV has

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