What is the general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c ?

The general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c is y=tan(c-1/2+t-1/2 t^2)

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solvefor y,arctan(y)+1/2 (1-t)^2=c

Solution

solve for

y, arctan (y) + \frac{1}{2} (1 - t)^{2} = c

Solution

y = tan (c - \frac{1}{2} + t - \frac{1}{2} t^{2})

Solution steps

arctan (y) + \frac{1}{2} (1 - t)^{2} = c

Move

\frac{1}{2} (1 - t)^{2}

to the right side

arctan (y) + \frac{1}{2} (1 - t)^{2} = c

Subtract

\frac{1}{2} (1 - t)^{2}

from both sides

arctan (y) + \frac{1}{2} (1 - t)^{2} - \frac{1}{2} (1 - t)^{2} = c - \frac{1}{2} (1 - t)^{2}

Simplify

arctan (y) + \frac{1}{2} (1 - t)^{2} - \frac{1}{2} (1 - t)^{2} = c - \frac{1}{2} (1 - t)^{2}

Simplify

arctan (y) + \frac{1}{2} (1 - t)^{2} - \frac{1}{2} (1 - t)^{2} : arctan (y)

arctan (y) + \frac{1}{2} (1 - t)^{2} - \frac{1}{2} (1 - t)^{2}

Add similar elements:

\frac{1}{2} (1 - t)^{2} - \frac{1}{2} (1 - t)^{2} = 0

= arctan (y)

Simplify

c - \frac{1}{2} (1 - t)^{2} : c - \frac{1}{2} + t - \frac{1}{2} t^{2}

c - \frac{1}{2} (1 - t)^{2}

(1 - t)^{2} : 1 - 2 t + t^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = t

= 1^{2} - 2 \cdot 1 \cdot t + t^{2}

Simplify

1^{2} - 2 \cdot 1 \cdot t + t^{2} : 1 - 2 t + t^{2}

1^{2} - 2 \cdot 1 \cdot t + t^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot t + t^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 t + t^{2}

= 1 - 2 t + t^{2}

= c - \frac{1}{2} (1 - 2 t + t^{2})

Expand

- \frac{1}{2} (1 - 2 t + t^{2}) : - \frac{1}{2} + t - \frac{1}{2} t^{2}

- \frac{1}{2} (1 - 2 t + t^{2})

Distribute parentheses

= (- \frac{1}{2}) \cdot 1 + (- \frac{1}{2}) (- 2 t) + (- \frac{1}{2}) t^{2}

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} t - \frac{1}{2} t^{2}

Simplify

- 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} t - \frac{1}{2} t^{2} : - \frac{1}{2} + t - \frac{1}{2} t^{2}

- 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} t - \frac{1}{2} t^{2}

1 \cdot \frac{1}{2} = \frac{1}{2}

1 \cdot \frac{1}{2}

Multiply:

1 \cdot \frac{1}{2} = \frac{1}{2}

= \frac{1}{2}

2 \cdot \frac{1}{2} t = t

2 \cdot \frac{1}{2} t

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2} t

Cancel the common factor:

2

= t \cdot 1

Multiply:

t \cdot 1 = t

= t

= - \frac{1}{2} + t - \frac{1}{2} t^{2}

= - \frac{1}{2} + t - \frac{1}{2} t^{2}

= c - \frac{1}{2} + t - \frac{1}{2} t^{2}

arctan (y) = c - \frac{1}{2} + t - \frac{1}{2} t^{2}

arctan (y) = c - \frac{1}{2} + t - \frac{1}{2} t^{2}

arctan (y) = c - \frac{1}{2} + t - \frac{1}{2} t^{2}

Apply trig inverse properties

arctan (y) = c - \frac{1}{2} + t - \frac{1}{2} t^{2}

arctan (x) = a \Rightarrow x = tan (a)

y = tan (c - \frac{1}{2} + t - \frac{1}{2} t^{2})

y = tan (c - \frac{1}{2} + t - \frac{1}{2} t^{2})

Popular Examples

(15-sqrt(3))/3 =5-2/3*csc(4x+pi/2)

\frac{15 - 3}{3} = 5 - \frac{2}{3} \cdot csc (4 x + \frac{π}{2})

tan(x)+4sin(x)=0

tan (x) + 4 sin (x) = 0

1+2*sin(x)=2

1 + 2 \cdot sin (x) = 2

1+2*sin(x)=3

1 + 2 \cdot sin (x) = 3

sin(α)=(120sin(60))/(1200)

sin (α) = \frac{120 sin ( 6 0 ^{\circ} )}{1200}

Frequently Asked Questions (FAQ)

What is the general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c ?
The general solution for solvefor y,arctan(y)+1/2 (1-t)^2=c is y=tan(c-1/2+t-1/2 t^2)