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).
-0.5736
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)
SOH CAH TOA is a way of remembering what the functions sin, cos, & tan mean in a right angle triangle. With a triangle with one of the acute angles labelled (theta) the longest side H (Hypotenuse), the side opposite the labelled angle O, and the short side closest to the angle A (Adjacent) SOH ->SIN(Theta)=0/H CAH ->COS(Theta)=A/H TOA ->TAN(Theta)=O/A
Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1
The identity for tan(theta) is sin(theta)/cos(theta).
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
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
If sin θ = tan θ, that means cos θ is 1 (since tan θ = (sin θ)/(cos θ)) (Usually in and equation a/b=a, b doesn't have to be 1 when a is 0, but cos θ = 1 if and only if sin θ = 0) The angles that satisfy cos θ = 1 is 2n(pi) (or 360n in degrees) When n is an integer. But if sin θ = tan θ = θ, the only answer is θ = 0. Because sin 0 is 0 and cos 0 is 1 and tan 0 is 0 The only answer would be when θ = 0.
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).
'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
sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = ±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).
0.75
[]=theta 1. sin[]=0.5sin[] Subtract 0.5sin[] from both sides.2. 0.5sin[]=0. Divide both sides by 0.5.3. Sin[] =0.[]=0 or pi (radians)
No, they cannot all be negative and retain the same value for theta, as is shown with the four quadrants and their trigonemtric properties. For example, in the first quadrant (0
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.