How to find a reference angle - ACT Math

Factor using the difference of two squares: _a_2 – _b_2 = (a + b)(a – b)

The identity 1 + tan2_Θ_ = sec2_Θ_ which can be rewritten as 1 = sec2_Θ_ – tan2_Θ_

So sec4_Θ_ – tan4_Θ_ = (sec2_Θ_ + tan2_Θ_)(sec2_Θ_ – tan2_Θ_) = (sec2_Θ_ + tan2_Θ_)(1) = sec2_Θ_ + tan2_Θ_

Convert cotΘ and secΘ to sinΘ and cosΘ and simplify the resulting complex fraction.

cotΘ = cosΘ secΘ = 1

sinΘ cosΘ

Factor the expression to get sinθ(cos2θ – 1).

The trig identity cos2θ + sin2θ = 1 can be reworked to becomes cos2θ – 1 = –sinθ resulting in the simplification of –sin3θ.

Cotθ can be written as cosθ/sinθ, which results in sinθ + cos2θ/sinθ.

Combining to get a single fraction results in (sin2θ + cos2θ)/sinθ.

Knowing that sin2θ + cos2θ = 1, we get 1/sinθ, which can be written as cscθ.

3600 – 3510 = 90

The first thing to do is to breakdown the meaning of each trig function, cot = cos/sin, sec = 1/cos, and sin = sin. Then put these back into the function and simplify if possible, so then (cos (Θ)/sin (Θ))*(1/cos (Θ))*(sin (Θ)) = (cos (Θ)*sin(Θ))/(sin (Θ)*cos(Θ)) = 1, since they all cancel out.

On the unit circle, (X,Y) = (cos Θ, sin Θ).

(cos Θ,sin Θ) = (cos 30º, sin 30º) = (√3 / 2 , 1 / 2)

At , sine and cosine have the same value.

Cotangent is given by .

Now we can evaluate the expression.

The reference angle is between and , starting on the positive -axis and rotating in a counter-clockwise manor.

To find the reference angle, we subtract for each complete revolution until we get a positive number less than .

is equal to two complete revolutions, plus a angle. Since is in Quadrant II, we subtract it from to get our reference angle:

A reference angle is the smallest possible angle between a given angle measurement and the x-axis.

In this case, our angle lies in Quadrant I, so the angle is its own reference angle.

Thus, the reference angle for is .

A reference angle is the smallest possible angle between a given angle measurement and the x-axis.

In this case, our angle lies in Quadrant III, so the angle is found by the formula .

Thus, the reference angle for is .

A reference angle is the smallest possible angle between a given angle measurement and the x-axis.

In this case, our angle lies in Quadrant II, so we can find our reference angle using the formula

.

Thus, the reference angle for is .

3600 – 3510 = 90

The first thing to do is to breakdown the meaning of each trig function, cot = cos/sin, sec = 1/cos, and sin = sin. Then put these back into the function and simplify if possible, so then (cos (Θ)/sin (Θ))*(1/cos (Θ))*(sin (Θ)) = (cos (Θ)*sin(Θ))/(sin (Θ)*cos(Θ)) = 1, since they all cancel out.

On the unit circle, (X,Y) = (cos Θ, sin Θ).

(cos Θ,sin Θ) = (cos 30º, sin 30º) = (√3 / 2 , 1 / 2)

Convert cotΘ and secΘ to sinΘ and cosΘ and simplify the resulting complex fraction.

cotΘ = cosΘ secΘ = 1

sinΘ cosΘ

Factor the expression to get sinθ(cos2θ – 1).

The trig identity cos2θ + sin2θ = 1 can be reworked to becomes cos2θ – 1 = –sinθ resulting in the simplification of –sin3θ.

Cotθ can be written as cosθ/sinθ, which results in sinθ + cos2θ/sinθ.

Combining to get a single fraction results in (sin2θ + cos2θ)/sinθ.

Knowing that sin2θ + cos2θ = 1, we get 1/sinθ, which can be written as cscθ.

Factor using the difference of two squares: _a_2 – _b_2 = (a + b)(a – b)

The identity 1 + tan2_Θ_ = sec2_Θ_ which can be rewritten as 1 = sec2_Θ_ – tan2_Θ_

So sec4_Θ_ – tan4_Θ_ = (sec2_Θ_ + tan2_Θ_)(sec2_Θ_ – tan2_Θ_) = (sec2_Θ_ + tan2_Θ_)(1) = sec2_Θ_ + tan2_Θ_

At , sine and cosine have the same value.

Cotangent is given by .

Now we can evaluate the expression.