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

The general solution for (1+cos^2(a))sin^2(a)=1 is a= pi/2+2pin,a=(3pi)/2+2pin

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

Solution

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

Solution

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

Degrees

a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (a) = u

- 1 + (2 - u^{2}) u^{2} = 0

- 1 + (2 - u^{2}) u^{2} = 0 : u = 1, u = - 1

- 1 + (2 - u^{2}) u^{2} = 0

Expand

- 1 + (2 - u^{2}) u^{2} : - 1 + 2 u^{2} - u^{4}

- 1 + (2 - u^{2}) u^{2}

= - 1 + u^{2} (2 - u^{2})

Expand

u^{2} (2 - u^{2}) : 2 u^{2} - u^{4}

u^{2} (2 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = u^{2}, b = 2, c = u^{2}

= u^{2} \cdot 2 - u^{2} u^{2}

= 2 u^{2} - u^{2} u^{2}

u^{2} u^{2} = u^{4}

u^{2} u^{2}

Apply exponent rule:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

= 2 u^{2} - u^{4}

= - 1 + 2 u^{2} - u^{4}

- 1 + 2 u^{2} - u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a = 0

- u^{4} + 2 u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

- v^{2} + 2 v - 1 = 0

Solve

- v^{2} + 2 v - 1 = 0 : v = 1

- v^{2} + 2 v - 1 = 0

Solve with the quadratic formula

- v^{2} + 2 v - 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 2, c = - 1

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

2^{2} - 4 (- 1) (- 1) = 0

2^{2} - 4 (- 1) (- 1)

Apply rule

- (- a) = a

= 2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- 2 \pm 0}{2 ( - 1 )}

v = \frac{- 2}{2 ( - 1 )}

\frac{- 2}{2 ( - 1 )} = 1

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = 1

The solution to the quadratic equation is:

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 1, u = - 1

Substitute back

u = sin (a)

sin (a) = 1, sin (a) = - 1

sin (a) = 1, sin (a) = - 1

sin (a) = 1 : a = \frac{π}{2} + 2 πn

sin (a) = 1

General solutions for

sin (a) = 1

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 = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

sin (a) = - 1 : a = \frac{3 π}{2} + 2 πn

sin (a) = - 1

General solutions for

sin (a) = - 1

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 = \frac{3 π}{2} + 2 πn

a = \frac{3 π}{2} + 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (1+cos^2(a))sin^2(a)=1 ?
The general solution for (1+cos^2(a))sin^2(a)=1 is a= pi/2+2pin,a=(3pi)/2+2pin