What is the general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) ?

The general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) is b=2pin,b=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(b)sin^2(b)=tan^2(b)+sin^2(b)

Solution

tan^{2} (b) sin^{2} (b) = tan^{2} (b) + sin^{2} (b)

Solution

b = 2 πn, b = π + 2 πn

Degrees

b = 0^{\circ} + 36 0^{\circ} n, b = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (b) sin^{2} (b) = tan^{2} (b) + sin^{2} (b)

Subtract

tan^{2} (b) + sin^{2} (b)

from both sides

tan^{2} (b) sin^{2} (b) - tan^{2} (b) - sin^{2} (b) = 0

Express with sin, cos

- sin^{2} (b) - tan^{2} (b) + sin^{2} (b) tan^{2} (b)

Use the basic trigonometric identity:

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

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

Simplify

- sin^{2} (b) - (\frac{sin ( b )}{cos ( b )})^{2} + sin^{2} (b) (\frac{sin ( b )}{cos ( b )})^{2} : \frac{- sin ^{2} ( b ) cos ^{2} ( b ) - sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

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

(\frac{sin ( b )}{cos ( b )})^{2} = \frac{sin ^{2} ( b )}{cos ^{2} ( b )}

(\frac{sin ( b )}{cos ( b )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( b )}{cos ^{2} ( b )}

sin^{2} (b) (\frac{sin ( b )}{cos ( b )})^{2} = \frac{sin ^{4} ( b )}{cos ^{2} ( b )}

sin^{2} (b) (\frac{sin ( b )}{cos ( b )})^{2}

(\frac{sin ( b )}{cos ( b )})^{2} = \frac{sin ^{2} ( b )}{cos ^{2} ( b )}

(\frac{sin ( b )}{cos ( b )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( b )}{cos ^{2} ( b )}

= \frac{sin ^{2} ( b )}{cos ^{2} ( b )} sin^{2} (b)

Multiply fractions:

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

= \frac{sin ^{2} ( b ) sin ^{2} ( b )}{cos ^{2} ( b )}

sin^{2} (b) sin^{2} (b) = sin^{4} (b)

sin^{2} (b) sin^{2} (b)

Apply exponent rule:

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

sin^{2} (b) sin^{2} (b) = sin^{2 + 2} (b)

= sin^{2 + 2} (b)

Add the numbers:

2 + 2 = 4

= sin^{4} (b)

= \frac{sin ^{4} ( b )}{cos ^{2} ( b )}

= - sin^{2} (b) - \frac{sin ^{2} ( b )}{cos ^{2} ( b )} + \frac{sin ^{4} ( b )}{cos ^{2} ( b )}

Combine the fractions

- \frac{sin ^{2} ( b )}{cos ^{2} ( b )} + \frac{sin ^{4} ( b )}{cos ^{2} ( b )} : \frac{- sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

Apply rule

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

= \frac{- sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

= - sin^{2} (b) + \frac{sin ^{4} ( b ) - sin ^{2} ( b )}{cos ^{2} ( b )}

Convert element to fraction:

sin^{2} (b) = \frac{sin ^{2} ( b ) cos ^{2} ( b )}{cos ^{2} ( b )}

= - \frac{sin ^{2} ( b ) cos ^{2} ( b )}{cos ^{2} ( b )} + \frac{- sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ^{2} ( b ) cos ^{2} ( b ) - sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

= \frac{- sin ^{2} ( b ) cos ^{2} ( b ) - sin ^{2} ( b ) + sin ^{4} ( b )}{cos ^{2} ( b )}

\frac{- sin ^{2} ( b ) + sin ^{4} ( b ) - cos ^{2} ( b ) sin ^{2} ( b )}{cos ^{2} ( b )} = 0

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

- sin^{2} (b) + sin^{4} (b) - cos^{2} (b) sin^{2} (b) = 0

Factor

- sin^{2} (b) + sin^{4} (b) - cos^{2} (b) sin^{2} (b) : - sin^{2} (b) (1 - sin^{2} (b) + cos^{2} (b))

- sin^{2} (b) + sin^{4} (b) - cos^{2} (b) sin^{2} (b)

Apply exponent rule:

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

sin^{4} (b) = sin^{2} (b) sin^{2} (b)

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

Factor out common term

- sin^{2} (b)

= - sin^{2} (b) (1 - sin^{2} (b) + cos^{2} (b))

- sin^{2} (b) (1 - sin^{2} (b) + cos^{2} (b)) = 0

Solving each part separately

sin^{2} (b) = 0 or 1 - sin^{2} (b) + cos^{2} (b) = 0

sin^{2} (b) = 0 : b = 2 πn, b = π + 2 πn

sin^{2} (b) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (b) = 0

General solutions for

sin (b) = 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}

b = 0 + 2 πn, b = π + 2 πn

b = 0 + 2 πn, b = π + 2 πn

Solve

b = 0 + 2 πn : b = 2 πn

b = 0 + 2 πn

0 + 2 πn = 2 πn

b = 2 πn

b = 2 πn, b = π + 2 πn

1 - sin^{2} (b) + cos^{2} (b) = 0 : b = \frac{π}{2} + πn

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

Rewrite using trig identities

1 - sin^{2} (b) + cos^{2} (b)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= 1 + cos (2 b)

1 + cos (2 b) = 0

Move

1

to the right side

1 + cos (2 b) = 0

Subtract

1

from both sides

1 + cos (2 b) - 1 = 0 - 1

Simplify

cos (2 b) = - 1

cos (2 b) = - 1

General solutions for

cos (2 b) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 b = π + 2 πn

2 b = π + 2 πn

Solve

2 b = π + 2 πn : b = \frac{π}{2} + πn

2 b = π + 2 πn

Divide both sides by

2

2 b = π + 2 πn

Divide both sides by

2

\frac{2 b}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

b = \frac{π}{2} + πn

b = \frac{π}{2} + πn

b = \frac{π}{2} + πn

Combine all the solutions

b = 2 πn, b = π + 2 πn, b = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

b = 2 πn, b = π + 2 πn

Popular Examples

cos(2θ)=(sqrt(3))/2 ,0<= θ<2pi

cos (2 θ) = \frac{3}{2}, 0 \leq θ < 2 π

7.5sin(x)=3.4cos(x)

7.5 sin (x) = 3.4 cos (x)

sec(θ)=8,tan(θ)=3sqrt(7)

sec (θ) = 8, tan (θ) = 37

tan(3x-4)=(-1)/2

tan (3 x - 4) = \frac{- 1}{2}

picos(2θ)=0

π cos (2 θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) ?
The general solution for tan^2(b)sin^2(b)=tan^2(b)+sin^2(b) is b=2pin,b=pi+2pin