0
It is a simple trigonometric equation. However, without information on whether the angles are measures in degrees or radians, and with no domain for theta, the equation cannot be solved.
Wiki User
∙ 12y ago
Add your answer:
Earn +20 pts
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
What is sec theta - 1 over sec theta?
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
What is sin theta cos theta?
It's 1/2 of sin(2 theta) .
What is the cot of a 45 degree angle?
1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1
What is cos theta and sin theta?
They are mathematical functions. Most people are introduced to them as trigonometric functions. In the context of a right angled triangle, with one of its angles being theta, Cos(theta) = The ratio of the lengths of the adjacent side and the hypotenuse. Sin(theta) = The ratio of the lengths of the opposite side and the hypotenuse. More advanced mathematicians will know them simply as the following infinite series: Cos(theta) = 1 - x2/2! + x4/4! - x6/6! + ... and Sin(theta) = x/1! - x3/3! + x5/5! - x7/7! + ... n! = 1*2*3* ... *n
How do you solve 4 cosine squared theta equals 1?
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
What is sec theta - 1 over sec theta?
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
What is sin theta cos theta?
It's 1/2 of sin(2 theta) .
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
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 verify 1 divided by cos to the second theta minus tan to the second theta equals cos to the second theta plus 1 divided by csc to the second theta?
2
Why is 2 sin theta cos theta equal to sin 2theta?
because sin(2x) = 2sin(x)cos(x)
What is sec squared theta times cos squared theta minus tan squared theta?
Tan^2
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
How do you calculate the area of anisosceles trapezoid if you only know the width of the base and the length and angle of the sides?
One way would be as follows: Let b represent the length of the base, l the length of each of the two sides, and theta the angle between the base and the two sides of length l. Now drop a perpendicular line from each vertex at the top of the trapezoid to the base. This yields two right triangles and a rectangle in the middle. The height of each right triangle (as well as the height of the rectangle) equals l*sin(theta) [because sin(theta)=opposite/hypotenuse] and the length of the base of each right triangle is l*cos(theta). The base of the rectangle is b minus the lengths of the two right triangles. Area of the trapezoid=2*area of each right triangle+area of the rectangle=2*(1/2)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=b*l*sin(theta)-l2*sin(theta)*cos(theta)
What is the cot of a 45 degree angle?
1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1
