What is the general solution for cos(θ)csc(θ)=sin(θ)sec(θ) ?

The general solution for cos(θ)csc(θ)=sin(θ)sec(θ) is θ= pi/4+pin,θ=(3pi)/4+pin

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cos(θ)csc(θ)=sin(θ)sec(θ)

Solution

cos (θ) csc (θ) = sin (θ) sec (θ)

Solution

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) csc (θ) = sin (θ) sec (θ)

Subtract

sin (θ) sec (θ)

from both sides

cos (θ) csc (θ) - sin (θ) sec (θ) = 0

Express with sin, cos

cos (θ) csc (θ) - sec (θ) sin (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= cos (θ) \frac{1}{sin ( θ )} - sec (θ) sin (θ)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ)

Simplify

cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ) : \frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ)

cos (θ) \frac{1}{sin ( θ )} = \frac{cos ( θ )}{sin ( θ )}

cos (θ) \frac{1}{sin ( θ )}

Multiply fractions:

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

= \frac{1 \cdot cos ( θ )}{sin ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

\frac{1}{cos ( θ )} sin (θ) = \frac{sin ( θ )}{cos ( θ )}

\frac{1}{cos ( θ )} sin (θ)

Multiply fractions:

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

= \frac{1 \cdot sin ( θ )}{cos ( θ )}

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

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

Least Common Multiplier of

sin (θ), cos (θ) : sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

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

sin (θ) cos (θ)

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos^{2} (θ) - sin^{2} (θ) = 0

Rewrite using trig identities

cos^{2} (θ) - sin^{2} (θ)

Use the Double Angle identity:

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

= cos (2 θ)

cos (2 θ) = 0

General solutions for

cos (2 θ) = 0

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

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

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

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

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Popular Examples

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64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

solvefor y,x=4sec(y)solve for

y, x = 4 sec (y)

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)csc(θ)=sin(θ)sec(θ) ?
The general solution for cos(θ)csc(θ)=sin(θ)sec(θ) is θ= pi/4+pin,θ=(3pi)/4+pin