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Popular Trigonometry >

((1+cos^2(a)))/(sin^2(a))= 5/3

Solution

\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}

Solution

a = \frac{π}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Degrees

a = 6 0^{\circ} + 36 0^{\circ} n, a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}

Subtract

\frac{5}{3}

from both sides

\frac{1 + cos ^{2} ( a )}{sin ^{2} ( a )} - \frac{5}{3} = 0

Simplify

\frac{1 + cos ^{2} ( a )}{sin ^{2} ( a )} - \frac{5}{3} : \frac{3 ( 1 + cos ^{2} ( a ) ) - 5 sin ^{2} ( a )}{3 sin ^{2} ( a )}

\frac{1 + cos ^{2} ( a )}{sin ^{2} ( a )} - \frac{5}{3}

Least Common Multiplier of

sin^{2} (a), 3 : 3 sin^{2} (a)

sin^{2} (a), 3

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin^{2} (a)

3

= 3 sin^{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

3 sin^{2} (a)

For

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

multiply the denominator and numerator by

3

\frac{1 + cos ^{2} ( a )}{sin ^{2} ( a )} = \frac{( 1 + cos ^{2} ( a ) ) \cdot 3}{sin ^{2} ( a ) \cdot 3}

For

\frac{5}{3} :

multiply the denominator and numerator by

sin^{2} (a)

\frac{5}{3} = \frac{5 sin ^{2} ( a )}{3 sin ^{2} ( a )}

= \frac{( 1 + cos ^{2} ( a ) ) \cdot 3}{sin ^{2} ( a ) \cdot 3} - \frac{5 sin ^{2} ( a )}{3 sin ^{2} ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{( 1 + cos ^{2} ( a ) ) \cdot 3 - 5 sin ^{2} ( a )}{3 sin ^{2} ( a )}

\frac{3 ( 1 + cos ^{2} ( a ) ) - 5 sin ^{2} ( a )}{3 sin ^{2} ( a )} = 0

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

3 (1 + cos^{2} (a)) - 5 sin^{2} (a) = 0

Rewrite using trig identities

(1 + cos^{2} (a)) \cdot 3 - 5 sin^{2} (a)

Use the Pythagorean identity:

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

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

= (1 + 1 - sin^{2} (a)) \cdot 3 - 5 sin^{2} (a)

Simplify

(1 + 1 - sin^{2} (a)) \cdot 3 - 5 sin^{2} (a) : - 8 sin^{2} (a) + 6

(1 + 1 - sin^{2} (a)) \cdot 3 - 5 sin^{2} (a)

Add the numbers:

1 + 1 = 2

= 3 (- sin^{2} (a) + 2) - 5 sin^{2} (a)

Expand

3 (- sin^{2} (a) + 2) : - 3 sin^{2} (a) + 6

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

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = - sin^{2} (a), c = 2

= 3 (- sin^{2} (a)) + 3 \cdot 2

Apply minus-plus rules

+ (- a) = - a

= - 3 sin^{2} (a) + 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= - 3 sin^{2} (a) + 6

= - 3 sin^{2} (a) + 6 - 5 sin^{2} (a)

Simplify

- 3 sin^{2} (a) + 6 - 5 sin^{2} (a) : - 8 sin^{2} (a) + 6

- 3 sin^{2} (a) + 6 - 5 sin^{2} (a)

Group like terms

= - 3 sin^{2} (a) - 5 sin^{2} (a) + 6

Add similar elements:

- 3 sin^{2} (a) - 5 sin^{2} (a) = - 8 sin^{2} (a)

= - 8 sin^{2} (a) + 6

= - 8 sin^{2} (a) + 6

= - 8 sin^{2} (a) + 6

6 - 8 sin^{2} (a) = 0

Solve by substitution

6 - 8 sin^{2} (a) = 0

Let:

sin (a) = u

6 - 8 u^{2} = 0

6 - 8 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

6 - 8 u^{2} = 0

Move

6

to the right side

6 - 8 u^{2} = 0

Subtract

6

from both sides

6 - 8 u^{2} - 6 = 0 - 6

Simplify

- 8 u^{2} = - 6

- 8 u^{2} = - 6

Divide both sides by

- 8

- 8 u^{2} = - 6

Divide both sides by

- 8

\frac{- 8 u ^{2}}{- 8} = \frac{- 6}{- 8}

Simplify

\frac{- 8 u ^{2}}{- 8} = \frac{- 6}{- 8}

Simplify

\frac{- 8 u ^{2}}{- 8} : u^{2}

\frac{- 8 u ^{2}}{- 8}

Apply the fraction rule:

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

= \frac{8 u ^{2}}{8}

Divide the numbers:

\frac{8}{8} = 1

= u^{2}

Simplify

\frac{- 6}{- 8} : \frac{3}{4}

\frac{- 6}{- 8}

Apply the fraction rule:

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

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = sin (a)

sin (a) = \frac{3}{2}, sin (a) = - \frac{3}{2}

sin (a) = \frac{3}{2}, sin (a) = - \frac{3}{2}

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

sin (a) = \frac{3}{2}

General solutions for

sin (a) = \frac{3}{2}

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}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn

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

sin (a) = - \frac{3}{2} : a = \frac{4 π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

sin (a) = - \frac{3}{2}

General solutions for

sin (a) = - \frac{3}{2}

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{4 π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

a = \frac{4 π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Combine all the solutions

a = \frac{π}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Graph

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