What is the general solution for sin^2(a)=((2tan(a)))/((1+tan^2(a))) ?

The general solution for sin^2(a)=((2tan(a)))/((1+tan^2(a))) is a=2pin,a=pi+2pin,a=1.10714…+pin

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Popular Trigonometry >

sin^2(a)=((2tan(a)))/((1+tan^2(a)))

Solution

sin^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 + tan ^{2} ( a ) )}

Solution

a = 2 πn, a = π + 2 πn, a = 1.10714 \dots + πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 + tan ^{2} ( a ) )}

Subtract

\frac{2 tan ( a )}{1 + tan ^{2} ( a )}

from both sides

sin^{2} (a) - \frac{2 tan ( a )}{1 + tan ^{2} ( a )} = 0

Simplify

sin^{2} (a) - \frac{2 tan ( a )}{1 + tan ^{2} ( a )} : \frac{sin ^{2} ( a ) ( 1 + tan ^{2} ( a ) ) - 2 tan ( a )}{1 + tan ^{2} ( a )}

sin^{2} (a) - \frac{2 tan ( a )}{1 + tan ^{2} ( a )}

Convert element to fraction:

sin^{2} (a) = \frac{sin ^{2} ( a ) ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

= \frac{sin ^{2} ( a ) ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )} - \frac{2 tan ( a )}{1 + tan ^{2} ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( a ) ( 1 + tan ^{2} ( a ) ) - 2 tan ( a )}{1 + tan ^{2} ( a )}

\frac{sin ^{2} ( a ) ( 1 + tan ^{2} ( a ) ) - 2 tan ( a )}{1 + tan ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (a) (1 + tan^{2} (a)) - 2 tan (a) = 0

Express with sin, cos

(1 + tan^{2} (a)) sin^{2} (a) - 2 tan (a)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= (1 + (\frac{sin ( a )}{cos ( a )})^{2}) sin^{2} (a) - 2 \cdot \frac{sin ( a )}{cos ( a )}

Simplify

(1 + (\frac{sin ( a )}{cos ( a )})^{2}) sin^{2} (a) - 2 \cdot \frac{sin ( a )}{cos ( a )} : \frac{sin ^{2} ( a ) ( cos ^{2} ( a ) + sin ^{2} ( a ) ) - 2 sin ( a ) cos ( a )}{cos ^{2} ( a )}

(1 + (\frac{sin ( a )}{cos ( a )})^{2}) sin^{2} (a) - 2 \cdot \frac{sin ( a )}{cos ( a )}

(1 + (\frac{sin ( a )}{cos ( a )})^{2}) sin^{2} (a) = \frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a)

(1 + (\frac{sin ( a )}{cos ( a )})^{2}) sin^{2} (a)

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= sin^{2} (a) (\frac{sin ^{2} ( a )}{cos ^{2} ( a )} + 1)

Join

1 + \frac{sin ^{2} ( a )}{cos ^{2} ( a )} : \frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )}

1 + \frac{sin ^{2} ( a )}{cos ^{2} ( a )}

Convert element to fraction:

1 = \frac{1 cos ^{2} ( a )}{cos ^{2} ( a )}

= \frac{1 \cdot cos ^{2} ( a )}{cos ^{2} ( a )} + \frac{sin ^{2} ( a )}{cos ^{2} ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )}

Multiply:

1 \cdot cos^{2} (a) = cos^{2} (a)

= \frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )}

= \frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a)

2 \cdot \frac{sin ( a )}{cos ( a )} = \frac{2 sin ( a )}{cos ( a )}

2 \cdot \frac{sin ( a )}{cos ( a )}

Multiply fractions:

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

= \frac{sin ( a ) \cdot 2}{cos ( a )}

= \frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a) - \frac{2 sin ( a )}{cos ( a )}

Multiply

\frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a) : \frac{sin ^{2} ( a ) ( cos ^{2} ( a ) + sin ^{2} ( a ) )}{cos ^{2} ( a )}

\frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a)

Multiply fractions:

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

= \frac{( cos ^{2} ( a ) + sin ^{2} ( a ) ) sin ^{2} ( a )}{cos ^{2} ( a )}

= \frac{( cos ^{2} ( a ) + sin ^{2} ( a ) ) sin ^{2} ( a )}{cos ^{2} ( a )} - \frac{sin ( a ) \cdot 2}{cos ( a )}

Least Common Multiplier of

cos^{2} (a), cos (a) : cos^{2} (a)

cos^{2} (a), cos (a)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos^{2} (a)

cos (a)

= cos^{2} (a)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos^{2} (a)

For

\frac{sin ( a ) \cdot 2}{cos ( a )} :

multiply the denominator and numerator by

cos (a)

\frac{sin ( a ) \cdot 2}{cos ( a )} = \frac{sin ( a ) \cdot 2 cos ( a )}{cos ( a ) cos ( a )} = \frac{sin ( a ) \cdot 2 cos ( a )}{cos ^{2} ( a )}

= \frac{( cos ^{2} ( a ) + sin ^{2} ( a ) ) sin ^{2} ( a )}{cos ^{2} ( a )} - \frac{sin ( a ) \cdot 2 cos ( a )}{cos ^{2} ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{( cos ^{2} ( a ) + sin ^{2} ( a ) ) sin ^{2} ( a ) - sin ( a ) \cdot 2 cos ( a )}{cos ^{2} ( a )}

= \frac{sin ^{2} ( a ) ( cos ^{2} ( a ) + sin ^{2} ( a ) ) - 2 sin ( a ) cos ( a )}{cos ^{2} ( a )}

\frac{( cos ^{2} ( a ) + sin ^{2} ( a ) ) sin ^{2} ( a ) - 2 cos ( a ) sin ( a )}{cos ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

(cos^{2} (a) + sin^{2} (a)) sin^{2} (a) - 2 cos (a) sin (a) = 0

Factor

(cos^{2} (a) + sin^{2} (a)) sin^{2} (a) - 2 cos (a) sin (a) : sin (a) (sin (a) (cos^{2} (a) + sin^{2} (a)) - 2 cos (a))

(cos^{2} (a) + sin^{2} (a)) sin^{2} (a) - 2 cos (a) sin (a)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (a) = sin (a) sin (a)

= (cos^{2} (a) + sin (a) sin (a)) sin (a) sin (a) - 2 cos (a) sin (a)

Factor out common term

sin (a)

= sin (a) ((cos^{2} (a) + sin^{2} (a)) sin (a) - 2 cos (a))

sin (a) (sin (a) (cos^{2} (a) + sin^{2} (a)) - 2 cos (a)) = 0

Rewrite using trig identities

sin (a) (sin (a) (cos^{2} (a) + sin^{2} (a)) - 2 cos (a))

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= sin (a) (- 2 cos (a) + sin (a) \cdot 1)

Simplify

sin (a) (- 2 cos (a) + sin (a) \cdot 1) : sin (a) (- 2 cos (a) + sin (a))

sin (a) (- 2 cos (a) + sin (a) \cdot 1)

Multiply:

sin (a) \cdot 1 = sin (a)

= sin (a) (sin (a) - 2 cos (a))

= sin (a) (- 2 cos (a) + sin (a))

sin (a) (- 2 cos (a) + sin (a)) = 0

Solving each part separately

sin (a) = 0 or - 2 cos (a) + sin (a) = 0

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

- 2 cos (a) + sin (a) = 0 : a = arctan (2) + πn

- 2 cos (a) + sin (a) = 0

Rewrite using trig identities

- 2 cos (a) + sin (a) = 0

Divide both sides by

cos (a), cos (a) \neq = 0

\frac{- 2 cos ( a ) + sin ( a )}{cos ( a )} = \frac{0}{cos ( a )}

Simplify

- 2 + \frac{sin ( a )}{cos ( a )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 2 + tan (a) = 0

- 2 + tan (a) = 0

Move

2

to the right side

- 2 + tan (a) = 0

Add

2

to both sides

- 2 + tan (a) + 2 = 0 + 2

Simplify

tan (a) = 2

tan (a) = 2

Apply trig inverse properties

tan (a) = 2

General solutions for

tan (a) = 2

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

a = arctan (2) + πn

a = arctan (2) + πn

Combine all the solutions

a = 2 πn, a = π + 2 πn, a = arctan (2) + πn

Show solutions in decimal form

a = 2 πn, a = π + 2 πn, a = 1.10714 \dots + πn

Graph

Popular Examples

solvefor a,sin^2(a)+cos^2(b)=1

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(a)=((2tan(a)))/((1+tan^2(a))) ?
The general solution for sin^2(a)=((2tan(a)))/((1+tan^2(a))) is a=2pin,a=pi+2pin,a=1.10714…+pin