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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.
