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Q: How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?

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Cos theta squared

1

The question contains an expression but not an equation. An expression cannot be solved.

Tan^2

You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.

cos2(theta) = 1 so cos(theta) = Â±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0

COS squared Theta + SIN squared Theta = 1; where Theta is the angles measurement in degrees.

The equation cannot be proved because of the scattered parts.

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)

If there is a plus in between, that would be equal to 1, as a result of the Pythagorean Theorem. Otherwise, you can convert this into other forms with some of the trigonometric identities for multiplication, but you won't really get it into a simpler form.

cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1

- cos theta

4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = Â±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

Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).

It is cotangent(theta).

sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)

(Sin theta + cos theta)^n= sin n theta + cos n theta

Zero. Anything minus itself is zero.

No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)

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

Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.

cosine (90- theta) = sine (theta)

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

zero