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Popular >

sin(θ)<0\land (csc(θ))(cos(θ))>0

Solution

sin (θ) < 0 and (csc (θ)) (cos (θ)) > 0

Solution

False for all θ \in R

Solution steps

sin (θ) < 0 and (csc (θ)) (cos (θ)) > 0

sin (θ) < 0 : - π + 2 πn < θ < 2 πn

sin (θ) < 0

For

sin (x) < a

, if

- 1 < a \leq 1

then

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (0) + 2 πn < θ < arcsin (0) + 2 πn

Simplify

- π - arcsin (0) : - π

- π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - 0

- π - 0 = - π

= - π

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

- π + 2 πn < θ < 0 + 2 πn

Simplify

- π + 2 πn < θ < 2 πn

csc (θ) cos (θ) > 0 : πn < θ < \frac{π}{2} + πn

csc (θ) cos (θ) > 0

Periodicity of

csc (θ) cos (θ) : π

csc (θ) cos (θ)

is composed of the following functions and periods:

csc (θ)

with periodicity of

2 π

The compound periodicity is:

= π

Express with sin, cos

csc (θ) cos (θ) > 0

Use the basic trigonometric identity:

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

\frac{1}{sin ( θ )} cos (θ) > 0

\frac{1}{sin ( θ )} cos (θ) > 0

Simplify

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

\frac{1}{sin ( θ )} cos (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{cos ( θ )}{sin ( θ )} > 0

Find the zeroes and undifined points of

\frac{cos ( θ )}{sin ( θ )}

for

0 \leq θ < π

To find the zeroes, set the inequality to zero

\frac{cos ( θ )}{sin ( θ )} = 0

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

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

Rewrite using trig identities

\frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

cot (θ) = 0

General solutions for

cot (θ) = 0

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Solutions for the range

0 \leq θ < π

θ = \frac{π}{2}

Find the undefined points:

θ = 0

Find the zeros of the denominator

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

0 \leq θ < π

θ = 0

0, \frac{π}{2}

Identify the intervals

0 < θ < \frac{π}{2}, \frac{π}{2} < θ < π

Summarize in a table:

cos (θ) sin (θ) \frac{c o s ( θ )}{s i n ( θ )} θ = 0 + 0 Undefined 0 < θ < \frac{π}{2} + + + θ = \frac{π}{2} 0 + 0 \frac{π}{2} < θ < π - + - θ = π - 0 Undefined

Identify the intervals that satisfy the required condition:

> 0

0 < θ < \frac{π}{2}

Apply the periodicity of

csc (θ) cos (θ)

πn < θ < \frac{π}{2} + πn

Combine the intervals

- π + 2 πn < θ < 2 πn and πn < θ < \frac{π}{2} + πn

Merge Overlapping Intervals

False for all θ \in R

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0<= sin^2(x)<= 1

0 \leq sin^{2} (x) \leq 1

cos(θ)=45\land 0<θ<90,sec(θ)

cos (θ) = 45 and 0^{\circ} < θ < 9 0^{\circ}, sec (θ)

sin(θ)<0\land cot(θ)<0

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5<= 20cos(pi/(20)(x-20))+23<= 20

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20