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# find tan 2a given that tangent A =4/3 and is in Quadrant 3

find tan 2A given:tangent A =4/3 and is in Quadrant 3

## Research, Knowledge and Information :

### *Double Angle Formula: If tanx= 3/4 and angle x is in ...

May 08, 2011 · If tanx= 3/4 and angle x is in quadrant 3 find tan 2x? ... Find Sin 2x, Cos 2x and Tan 2x from the given information - Duration: ... (2A), Cos(2A), Tan(2A ...

### 5.4 Multiple angle and half angle identities - jwbales.us

... is in quadrant II and $$\sin A = \frac{2}{3}$$ find $$\sin ( 2A )$$ and \ ... Given that $$\tan A = -3$$, find $$\tan ( 2A)$$ ... Tangent half-angle identities.

### Trigonometric Functions - Questions With Answers

Solve trigonometry Questions related to trigonometric functions. ... Given that sin (x) = 1 / 4, ... Questions 3: tan(x) = 4 and x is in quadrant III. Find the exact ...

### given cos x=-3/4, with x in quadrant 2. find the remaining ...

given cos x=-3/4, with x in quadrant 2. find the remaining five trigonometric functions of x. ... prove that sec^2a+cos^2a can never be less than 2 a telephone ...

### Given tantheta=-3/4 and 90<theta<180, how do you find tan ...

... +- d/(2a) = 8/6 +- 10/6 = (4 +- 5)/3 x1 = 3, and x2 = - 1/3 tan (t/2) = x1 = 3 and tan (t/2) = x2 = - 1/3 Because t is in Quadrant II ... Given #tantheta=-3/4 ...

### If cos A=-4/5 for angle A in Quadrant 2, find tan 2A.?

If cos A=-4/5 for angle A in Quadrant 2, find tan 2A.? ... given that tan(θ ... If sinA 4 5 and angle A is in Quadrant I, cosA 3 Read more. Positive: 66 %. Quadrants ...

### Given that tan A = 3/4, cos B = -(3/5) and A and B are in the ...

Note that tan(A) is 3/4, ... Given that tan A = 3/4, cos B = -(3/5) and A and B are in the same quadrant, can you evaluate tan B + cos ...

### Given that tan 2A = -8/15. find tan A? - Weknowtheanswer

Given that tan 2A = -8/15. find ... 0If tan 2A = cot ( A ... Given tan A =4/3, find the other ... given that cos a = -?39/8, a in quadrant 2, find the exact value of ...

## Suggested Questions And Answer :

### find tan 2a given that tangent A =4/3 and is in Quadrant 3

tan(x)=4/3, yuze invers-tan tu get x x=53.13010235415591 deg 2x=106.26020470831182 tan(106.26...)=-3.428571428571461

### How do I find out sin(A+B) and cos(A+B)??

Question: Given: A and B are in quadrant 4, cos A=(15/17) and tan B=-(5/12); How do I find out sin(A+B) and cos(A+B)?? In the fourth quadrant, only cos() is positive. Both sin() and tan() are negative. cos(A) = 15/17. Now draw a right angled triangle, with the hypotenuse = 17 and the adjacent side = 15 (that will give you cos() = 15/17 = cos(A) ) The opposite side then = sqrt(17^2 - 15^2) = sqrt(64) = 8 Therefore sin() = opposite/hypotenuse = 8/17. But A is in the fourth quadrant making sin(A) negative. i.e. sin(A) = -8/17   tan(B) = -5/12. Now draw a right angled triangle, with the opposite side = 5 and the adjacent side = 12 (that will give you tan() = 5/12 = tan(B).) (ignoring sign for the moment) The hypotenuse then = sqrt(5^2 + 12^2) = sqrt(169) = 13. Therefore sin() = opposite/hypotenuse = 5/13. But B is in the fourth quadrant making sin(B) negative. i.e. sin(B) = -5/13 Also cos() = adjacent/hypotenuse = 12/13. Again, B is in the fourth quadrant, this time making cos(B) positive. i.e. cos(B) = 12/13   Now we work out sin(A+B) and cos(A+B). sin(A+B) = sin(A).cos(B) + cos(A).sin(B) sin(A+B) = (-8/17).(12/13) + (15/17).(-5/13) = -96/221 - 75/221 = -171/221 sin(A+B) =171/221 cos(A+B) = cos(A).cos(B) - sin(A).sin(B) cos(A+B) = (15/17).(12/13) - (-8/17).(-5/13) = 180/221 - 40/221 = 140/221 cos(A+B) = 140/221

look like tangent kervs, but offset bi -90deg  (pi/2)...tan(theta+90) ?? horizontal distans tween 2 vertikal lines=1 ?? that meen angel vary bi 180 deg (pi) as x vary bi 1 impli tan[(theta+90)*pi] B...yu sed vertikal lines be distans=1=period av tan...same as angel vary bi pi C...koef av x???...tan(angel*pi) D faze shift=0.5 in distans or pi/2 in angel E???? F ????? "korrekt" ???? yu want it RITE ??

### How do find cos(alpha), tan(alpha), sin(beta), and tan(beta)?

if sine(alfa)=5/13, take the INVERS sine(5/13) =invers sine(0.38461538461538463) =22.6198649476 deg But if yu want quadrant 2, yu want 180 deg -22.619865 deg =157.3801351 deg

### line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them There are two lines that satisfy the given parameters, one above and to the left of line d, and one below and to the right of line d. To find a point 6 units away from line d, we need a line that is perpendicular to d. We get the equation of that line by using the negative reciprocal of the m component, 2: that gives us -1/2. The equation of this line is y = -1/2 x + 4, using the same y intercept. A segment of that line that is 6 units represents the hypotenuse of a right triangle. If we are looking at the segment that extends to the left of the given y intercept, we consider the distance to be negative (-6). If we are looking at the segment that extends to the right of the given y intercept, we consider the distance to be positive (+6). To find the x and y co-ordinates, we use the sine and the cosine of the angle (we'll call it a) formed by this new line and the x axis. The slope m of a line is the y distance divided by the x distance. That is also the definition of the tangent of the angle. Using the inverse tangent, we determine that the angle is tan^-1 (-1/2) =  -26.565 degrees. We need to keep all the signs straight in order to get the correct values. The sine of -26.565 degrees is -0.4472. The cosine of -26.565 degrees is 0.8944. y1 / -6 = sin a y1 = -6 sin a = -6 * -0.4472 = 2.6832 This is measured from a horizontal line through the y intercept, because we are constructing a right-triangle with one corner at the y intercept, so the point's y co-ordinate is actually y = 2.6832 + 4 = 6.6832 x / -6 = cos a x = -6 * 0.8944 = -5.3664 Making a quick check using x^2 + y^2 = r^2: (-5.3664)^2 + 2.6832^2 = 28.79825 + 7.19956 = 35.99781   -6^2 = 36 Close enough considering the rounding in the calculations. We have co-ordinate (-5.3664, 6.6832) lying 6 units from line d. Substituting those values into y = 2x + c we have 6.6832 = 2 * (-5.3664) + c 6.6832 = -10.7328 + c 6.6832 + 10.7328 = c = 17.416 The equation of a line 6 units away from d and parallel to it, located above it, is y = 2x + 17.416    <<<<<<<< Working in the other direction, on the line below and to the right of line d, y2 / 6 = sin a y2 = 6 * sin a = 6 * -0.4472 = -2.6832 (that is, 2.6832 below d's y intercept) y = 4 - 2.6832 = 1.3168 x / 6 = cos a x = 6 * cos a = 6 * 0.8944 = 5.3664 The co-ordinates for this point, (5.3664, 1.3168), when substituted into the equation y = 2x + f (f is a new y intercept) gives 1.3168 = 2 * 5.3664 + f 1.3168 - 10.7328 = f = -9.416 The equation of a line 6 units away from d and parallel to it, located below it, is y = 2x - 9.416    <<<<<<<<

### Given tanA = -2/3, A is in quadrant II, find Sin2A

If tanA=-2/3 we can use a right-angled triangle to find sinA and cosA, then use 2sinAcosA=sin2A. Hypotenuse=sqrt(2^2+3^2)=sqrt(13). sinA=2/sqrt(13), cosA=-3/sqrt(13), for A in the second quadrant, where sine is positive and cosine and tangent are negative, so sin2A=2(2/sqrt(13))(-3/sqrt(13))=-12/13. The angle A is in the second quadrant and has a value of about 146 degrees. 2A is therefore about 292 degrees in the fourth quadrant.

### The equation of the circle is x^2+y^2+6x+2y-6=0 Find the exact values of k for which y=x+k is a tangent to the circle.

The equation of the circle can be written (x+3)^2+(y+1)^2=16 by completing squares for x and y. From this (y+1)^2=16-(x+3)^2. The centre of the circle is at (-3,-1) and the radius is 4. Differentiating wrt x we get: 2(x+3)+2(y+1)y'=0 so y'=-(x+3)/(y+1). This is the gradient at point (x,y), so since the gradient of the tangent has to be 1, because y=x+k has a gradient of 1, y'=1. If the point where the tangent line meets the circle is (p,q) then the tangent line is y-q=x-p so y=x+q-p and k=q-p. -(p+3)/√(16-(p+3)^2)=1, being the gradient at (p,q) or (p,√(16-(p+3)^2)). Therefore -(p+3)=√(16-(p+3)^2); squaring both sides we have: (p+3)^2=16-(p+3)^2 and (p+3)^2=8, p=-3±2√2. From the standardised equation of the circle (q+1)^2=16-8=8 and q=-1±2√2. There are two points on the circle where the tangent line has a slope of 1 so there are two possible values for k (see picture). These points are diametrically opposite. The first point is when x is negative and y is positive (2nd quadrant) and the opposite point is when x is negative and y is negative (3rd quadrant). The points are (-3-2√2,-1+2√2) and (-3+2√2,-1-2√2), giving values of k=2+4√2, 2-4√2. Note that the line y+1=x+3, that is, y=x+2 is the line passing through the circle's centre parallel to y=x+k. This line is equidistant from the two lines y=x+2+4√2 and y=x+2-4√2. From this fact it is possible to calculate the equations of the lines y=x+k for the two values of k without resorting to calculus. The two tangent points are radially distant from the centre by 4, at opposite ends of the diameter. The radius is the hypotenuse of an isosceles right-angled triangle, so the two sides have length a, given by 2a^2=4^2, making a=√8=2√2. The points on the circumference can then be calculated: y=-3±2√2 and x=-1±2√2. By plugging in the x-y values into y=x+k, or using slope-intercept form, the two values of k can be calculated: y+1-2√2=x+3+2√2, y=2+4√2; y+1+2√2=x+3-2√2, y=2-4√2.

### find an equation for the line tangent to curve at the point defined by the given value of t

When t=3(pi)/4, x=3sin(3(pi)/4)=3sin((pi)/4)=3sqrt(2)/2, and y=3cos(3(pi)/4)=-3sqrt(2)/2. dy/dx*dx/dt=dy/dt, so dy/dx*3cos(t)=-3sint; dy/dx=-tan(t). This is the tangent, and at t=3(pi)/4 dy/dx=-1. The equation of the tangent has the form y=mx+c, where m=dy/dx=-1. To find c we substitute (3sqrt(2)/2,-3sqrt(2)/2) into the equation: -3sqrt(2)/2=-3sqrt(2)/2+c, from which c=0, and y=-x is the equation of the tangent when t=3(pi)/4.

### how do find the size of sides in an octogon, when nothing is given?

how do find the size of sides in an octogon, when nothing is given? the size of the piece is 2.250x2.250. There are eight vertices in an octagon, so there are eight internal angles from the center to the vertices. Each angle is three hundred sixty degrees divided by eight, or forty-five degrees. We can use the tangent of half that angle to find the length of one half of a side of the octagon (as shown in the figure). The base of the angle is one half of the square that encloses the octagon. x / 1.125 = tan 22.5 x / 1.125 = 0.414213 x = 0.414213 * 1.125 x = 0.46599 The length of the side is twice that. s = 2 * x s = 2 * 0.46599 s = 0.93198