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Popular Trigonometry >

csc(θ)<0\land cos(θ)>0

Solution

csc (θ) < 0 and cos (θ) > 0

Solution

False for all θ \in R

Solution steps

csc (θ) < 0 and cos (θ) > 0

csc (θ) < 0 :

False for all

θ \in R

csc (θ) < 0

Express with sin, cos

csc (θ) < 0

Use the basic trigonometric identity:

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

\frac{1}{sin ( θ )} < 0

\frac{1}{sin ( θ )} < 0

If

\frac{1}{a} < 0

then

a < 0

sin (θ) < 0

No Solution for θ \in R

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

cos (θ) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < θ < arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

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

Combine the intervals

False for all θ \in R and - \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

Merge Overlapping Intervals

False for all θ \in R

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