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Almost by definition, tan θ = sin θ / cos θ
You can convert this to sine θ in several ways, for example:
sin θ / cos θ = sin θ / cos (pi/2 - θ)
Or here is another way, using the Pythagorean identity:
sin θ / cos θ = sin θ / root(1 - sin2θ)
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What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
What is tan theta in terms of sin theta in quadrant II?
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
How do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
How do you simplify bracket 1 plus tan theta bracket bracket 5 sin theta -2 bracket equals 0?
(in a past paper it asks u to solve this for -180</=theta<180, so I have solved it) Tan theta =-1, so theta = -45. Use CAST diagram to find other values of theta for -180</=theta<180: Theta (in terms of tan) = -ve, other value is in either S or C. But because of boundaries value can only be in S. So other value= 180-45=135. Do the same for sin. Sin theta=2/5 so theta=23.6 CAST diagram, other value in S because theta (in terms of sin)=+ve. So other value=180-23.6=156.4.
Why tan theta sin theta divided by cos theta?
The answer depends on how the ratios are defined. In some cases tan is DEFINED as the ratio of sine and cosine rather than from the angle in a right angled triangle.If the trig ratios were defined in terms of a right angled triangle, thensine is the ratio of the opposite side to the hypotenuse,cosine is the ratio of the adjacent side to the hypotenuse,and tangent is the ratio of the opposite side to the adjacent side.It is then easy to see that sin/cos = (opp/hyp)/(adj/hyp) = opp/adj = tan.If sine and cosine are defined as infinite sums for angles measured in radians, iesin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...andcos = 1 - x^2/2! + x^4/4! - x^6/6! + ...then it is less easy to see tan = sin/cos.
How do you simplify sin theta times csc theta divided by tan theta?
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).
What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
How do you simplify tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
Sin theta 0 and tan theta 0?
4
What is tan theta in terms of sin theta in quadrant II?
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
How do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
How do you simplify bracket 1 plus tan theta bracket bracket 5 sin theta -2 bracket equals 0?
(in a past paper it asks u to solve this for -180</=theta<180, so I have solved it) Tan theta =-1, so theta = -45. Use CAST diagram to find other values of theta for -180</=theta<180: Theta (in terms of tan) = -ve, other value is in either S or C. But because of boundaries value can only be in S. So other value= 180-45=135. Do the same for sin. Sin theta=2/5 so theta=23.6 CAST diagram, other value in S because theta (in terms of sin)=+ve. So other value=180-23.6=156.4.
If sin theta equals 3/4 and theta is in quadrant II what is the value of tan theta?
0.75
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
How do you work out the remaining side of a right angled triangle using theta and one side?
Suppose triangle ABC is right angled at C. Suppose you are given that the angle at B is theta. Thenif you know the length of AB (the hypotenuse), thenBC = AB*cos(theta) andAC = AB*sin(theta)if you know the length of BC, thenAB = BC/cos(theta) andAC = BC*tan(theta)if you know the length of AC, thenAB= AC/sin(theta) andBC = AC/tan(theta)
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
Why tan theta sin theta divided by cos theta?
The answer depends on how the ratios are defined. In some cases tan is DEFINED as the ratio of sine and cosine rather than from the angle in a right angled triangle.If the trig ratios were defined in terms of a right angled triangle, thensine is the ratio of the opposite side to the hypotenuse,cosine is the ratio of the adjacent side to the hypotenuse,and tangent is the ratio of the opposite side to the adjacent side.It is then easy to see that sin/cos = (opp/hyp)/(adj/hyp) = opp/adj = tan.If sine and cosine are defined as infinite sums for angles measured in radians, iesin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...andcos = 1 - x^2/2! + x^4/4! - x^6/6! + ...then it is less easy to see tan = sin/cos.
