What is the general solution for (1-cos(a))(1+cos(a))=tan(a)sin(a) ?

The general solution for (1-cos(a))(1+cos(a))=tan(a)sin(a) is a=2pin,a=pi+2pin

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Popular >

(1-cos(a))(1+cos(a))=tan(a)sin(a)

Solution

(1 - cos (a)) (1 + cos (a)) = tan (a) sin (a)

Solution

a = 2 πn, a = π + 2 πn

Degrees

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

Solution steps

(1 - cos (a)) (1 + cos (a)) = tan (a) sin (a)

Subtract

tan (a) sin (a)

from both sides

(1 - cos (a)) (1 + cos (a)) - tan (a) sin (a) = 0

Rewrite using trig identities

(1 + cos (a)) (1 - cos (a)) - sin (a) tan (a)

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

1 - cos (a)

Multiply by

\frac{1 + cos ( a )}{1 + cos ( a )}

= \frac{( 1 - cos ( a ) ) ( 1 + cos ( a ) )}{1 + cos ( a )}

Apply Difference of Two Squares Formula:

a^{2} - y^{2} = (a + y) (a - y)

(1 - cos (a)) (1 + cos (a)) = 1 - cos^{2} (a)

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

Use the Pythagorean identity:

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

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

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

1 + cos (a)

= sin^{2} (a)

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

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

Factor

sin^{2} (a) - sin (a) tan (a) : sin (a) (sin (a) - tan (a))

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

Apply exponent rule:

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

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

= sin (a) sin (a) - sin (a) tan (a)

Factor out common term

sin (a)

= sin (a) (sin (a) - tan (a))

sin (a) (sin (a) - tan (a)) = 0

Solving each part separately

sin (a) = 0 or sin (a) - tan (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

sin (a) - tan (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) - tan (a) = 0

Express with sin, cos

sin (a) - tan (a)

Use the basic trigonometric identity:

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

= sin (a) - \frac{sin ( a )}{cos ( a )}

Simplify

sin (a) - \frac{sin ( a )}{cos ( a )} : \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

sin (a) - \frac{sin ( a )}{cos ( a )}

Convert element to fraction:

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

= \frac{sin ( a ) cos ( a )}{cos ( a )} - \frac{sin ( a )}{cos ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

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

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

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

Factor

- sin (a) + cos (a) sin (a) : sin (a) (cos (a) - 1)

- sin (a) + cos (a) sin (a)

Factor out common term

sin (a)

= sin (a) (- 1 + cos (a))

sin (a) (cos (a) - 1) = 0

Solving each part separately

sin (a) = 0 or cos (a) - 1 = 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

cos (a) - 1 = 0 : a = 2 πn

cos (a) - 1 = 0

Move

1

to the right side

cos (a) - 1 = 0

Add

1

to both sides

cos (a) - 1 + 1 = 0 + 1

Simplify

cos (a) = 1

cos (a) = 1

General solutions for

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 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

Combine all the solutions

a = 2 πn, a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (1-cos(a))(1+cos(a))=tan(a)sin(a) ?
The general solution for (1-cos(a))(1+cos(a))=tan(a)sin(a) is a=2pin,a=pi+2pin