It's easiest to show all of the work (explanations/identities), and x represents theta.
cosxcotx + sinx = cscx
cosx times cosx/sinx + sinx = csc x (Quotient Identity)
cosx2 /sinx + sinx = csc x (multiplied)
1-sinx2/sinx + sinx = csc x (Pythagorean Identity)
1/sinx - sinx2/sinx + sinx = csc x (seperate fraction)
1/sinx -sinx + sinx = csc x (canceled)
1/sinx = csc x (cancelled)
csc x =csc x (Reciprocal Identity)
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
To integrate ( \cos^2 \theta \sin \theta ), you can use a substitution method. Let ( u = \cos \theta ), then ( du = -\sin \theta , d\theta ). The integral becomes ( -\int u^2 , du ), which evaluates to ( -\frac{u^3}{3} + C ). Substituting back, the final result is ( -\frac{\cos^3 \theta}{3} + C ).
sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.
You must think of the unit circle. negative theta is in either radians or degrees and represents a specific area on the unit circle. Remember the unit circle is also like a coordinate plane and cos is the x while sin is the y coordinate. Here is an example: cos(-45): The cos of negative 45 degrees is pi/4 and cos(45) is also pi/4
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
No.
The equation that satisfies the condition "what divided by cosine squared theta equals one" is simply the expression itself. If we let ( x ) be the quantity, then the equation can be expressed as ( \frac{x}{\cos^2 \theta} = 1 ). Solving for ( x ) gives ( x = \cos^2 \theta ). Thus, ( \cos^2 \theta ) divided by ( \cos^2 \theta ) equals one.
The question contains an expression but not an equation. An expression cannot be solved.
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
Cos theta squared
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
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.
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89