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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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∙ 10y ago
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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 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.
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
Why tan theta sin theta divided by cos theta?
The expression tan(theta) sin(theta) / cos(theta) simplifies to sin^2(theta) / cos(theta). In trigonometry, sin^2(theta) is equal to (1 - cos^2(theta)), so the expression can be further simplified to (1 - cos^2(theta)) / cos(theta).
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 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.
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
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 expression tan(theta) sin(theta) / cos(theta) simplifies to sin^2(theta) / cos(theta). In trigonometry, sin^2(theta) is equal to (1 - cos^2(theta)), so the expression can be further simplified to (1 - cos^2(theta)) / cos(theta).
