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Cot 70 plus 4cos70 equals?
cot 70 + 4 cos 70 = cos 70 / sin 70 + 4 cos 70 = cos 70 (1/sin 70 + 4) = cos 70 (csc 70 + 4) Numerical answer varies, depending on whether 70 is in degrees, radians, or grads.
When does cotangent equal -1?
cot[x]= -1 cot[x] = cos[x] / sin[x] cos[x] / sin[x] = -1 cos[x] = -sin[x] |cos[x]| = |sin[x]| at every multiple of Pi/4 + Pi/2. However, the signs disagree at 3Pi/4 + nPi, where n is an integer.
How tan9-tan27-tan63 tan81 equals 4?
tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
How do you confirm this trig identity?
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
Find the equation of the line tangent to the graph of y equals cos2x at x equals pi divided by 4?
0.5
How was it determined that the tangent of pi divided by 4 is 1?
As tan(x)=sin(x)/cos(x) and sin(pi/4) = cos(pi/4) (= sqrt(2)/2) then tan(pi/4) = 1
How does the graph of the cosine function differ from a graph of a sine function?
the graph of cos(x)=1 when x=0the graph of sin(x)=0 when x=0.But that only tells part of the story. The two graphs are out of sync by pi/2 radians (or 90°; also referred to as 1/4 wavelength or 1/4 cycle). One cycle is 2*pi radians (the distance for the graph to get back where it started and repeat itself.The cosine graph is 'ahead' (leads) of the sine graph by 1/4 cycle. Or you can say that the sine graph lags the cosine graph by 1/4 cycle.
How do you graph cos x plus pie divided by 4?
amplitude=1 period=2 pi phase shift=0 vertical translation=pi/4 it will start at (0,1) on the y-axis and cross through pi on the x-axis, then the min will be in the middle of pi and 2 pi at -1. there will be one complete wave to 2 pi on your graph (one curve on top and one on bottom of the x-axis), but then you need to shift it to the left, so that the graph will start at (0,-1).
What is the exact value of cos theta if csc theta -4 with theta in quadrant III?
csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4
This graph is excellent for showing business cycles?
Unfortunately the graph does not show.. But, i can tell you that business cycle is divided into: 1) introduction - start of the graph 2) growth - graph goes up 3) maturity - graph is static and slowly pointing doen 4)decline - graph starts to go down.. if your graph is this way, then the answer is yes..
What does Sin squared x Cos squared x equal?
sin2 x = (1/2)(1 - cos 2x) cos2 x = (1/2)(1 + cos 2x) Multiplying both you get (1/4) (1 - cos2 2x) Which is equal to (1/4) (1 - (1/2) (1 + cos 4x) = (1/8) (2 - 1 - cos 4x) = (1/8) (1 - cos 4x) Or If it is the trigonomic function, sin squared x and cosine squared x is equal to one
How do you proof that there is one and only one value of theta makes cos of theta equals 1 true?
You cannot prove it because it is not true! cos(0) = 1 cos(2*pi) = 1 cos(4*pi) = 1 ... cos(2*k*pi) = 1 for all integers k or, if you still work in degrees, cos(0) = 1 cos(360) = 1 cos(720) = 1 ... cos(k*360) = 1 for all integers k
What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
Cos 195 degree use the trigonometry functions of quadrantal angles and special to find the exact value?
cos(195) = cos(180 + 15) = cos(180)*cos(15) - sin(180)*sin(15) = -1*cos(15) - 0*sim(15) = -cos(15) = -cos(60 - 45) = -[cos(60)*cos(45) + sin(60)*sin(45)] = -(1/2)*sqrt(2)/2 - sqrt(3)/2*sqrt(2)/2 = - 1/4*sqrt(2)*(1 + sqrt3) or -1/4*[sqrt(2) + sqrt(6)]
What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
Cot 70 plus 4cos70 equals?
cot 70 + 4 cos 70 = cos 70 / sin 70 + 4 cos 70 = cos 70 (1/sin 70 + 4) = cos 70 (csc 70 + 4) Numerical answer varies, depending on whether 70 is in degrees, radians, or grads.
