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)
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
The question contains an expression but not an equation. An expression cannot be solved.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
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
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.
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
Verify the identity:1/(cos θ)^2 - (tan θ)^2 = (cos θ)^2 + 1/(csc θ)^21/(cos θ)^2 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?1 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?(cos θ)^2/(cos θ)^2 = 1 ?1 = 1 TrueMethod 21/(cos θ)2 - (tan θ)2 =? (cos θ)2 + 1/(cscθ)21/(cos θ)2-(sinθ)2/(cos θ)2=? (cosθ)2+ sin(θ)21/(cos θ)2[1-sin(θ)2]=? cos(θ)2+sin(θ)21/cos(θ)2(cos(θ)2)=? 11=1 True
(Sin theta + cos theta)^n= sin n theta + cos n theta
It is cotangent(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
Zero. Anything minus itself is zero.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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
'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
cosine (90- theta) = sine (theta)
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
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.
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
It's 1/2 of sin(2 theta) .
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
The equation cannot be proved because of the scattered parts.
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