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Popular >

sin(a)tan(a)>0

Solution

sin (a) tan (a) > 0

Solution

2 πn < a < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < a < 2 π + 2 πn

Interval Notation

(2 πn, \frac{π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Decimal Notation

2 πn < a < 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn < a < 6.28318 \dots + 2 πn

Solution steps

sin (a) tan (a) > 0

Periodicity of

sin (a) tan (a) : 2 π

sin (a) tan (a)

is composed of the following functions and periods:

sin (a)

with periodicity of

2 π

The compound periodicity is:

= 2 π

Express with sin, cos

sin (a) tan (a) > 0

Use the basic trigonometric identity:

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

sin (a) \frac{sin ( a )}{cos ( a )} > 0

sin (a) \frac{sin ( a )}{cos ( a )} > 0

Simplify

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

sin (a) \frac{sin ( a )}{cos ( a )}

Multiply fractions:

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

= \frac{sin ( a ) sin ( a )}{cos ( a )}

sin (a) sin (a) = sin^{2} (a)

sin (a) sin (a)

Apply exponent rule:

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

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

= sin^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= sin^{2} (a)

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

\frac{sin ^{2} ( a )}{cos ( a )} > 0

Find the zeroes and undifined points of

\frac{sin ^{2} ( a )}{cos ( a )}

for

0 \leq a < 2 π

To find the zeroes, set the inequality to zero

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

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

\frac{sin ^{2} ( a )}{cos ( a )} = 0, 0 \leq a < 2 π

Rewrite using trig identities

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

Divide both sides by

cos (a), cos (a) \neq = 0

\frac{\frac{s i n ^{2} ( a )}{c o s ( a )}}{cos ( a )} = \frac{0}{cos ( a )}

Simplify

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

Use the basic trigonometric identity:

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

tan^{2} (a) = 0

tan^{2} (a) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (a) = 0

General solutions for

tan (a) = 0

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

a = 0 + πn

a = 0 + πn

Solve

a = 0 + πn : a = πn

a = 0 + πn

0 + πn = πn

a = πn

a = πn

Solutions for the range

0 \leq a < 2 π

a = 0, a = π

Find the undefined points:

a = \frac{π}{2}, a = \frac{3 π}{2}

Find the zeros of the denominator

cos (a) = 0

General solutions for

cos (a) = 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}

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

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

Solutions for the range

0 \leq a < 2 π

a = \frac{π}{2}, a = \frac{3 π}{2}

0, \frac{π}{2}, π, \frac{3 π}{2}

Identify the intervals

0 < a < \frac{π}{2}, \frac{π}{2} < a < π, π < a < \frac{3 π}{2}, \frac{3 π}{2} < a < 2 π

Summarize in a table:

sin^{2} (a) cos (a) \frac{s i n ^{2} ( a )}{c o s ( a )} a = 0 0 + 0 0 < a < \frac{π}{2} + + + a = \frac{π}{2} + 0 Undefined \frac{π}{2} < a < π + - - a = π 0 - 0 π < a < \frac{3 π}{2} + - - a = \frac{3 π}{2} + 0 Undefined \frac{3 π}{2} < a < 2 π + + + a = 2 π 0 + 0

Identify the intervals that satisfy the required condition:

> 0

0 < a < \frac{π}{2} or \frac{3 π}{2} < a < 2 π

Apply the periodicity of

sin (a) tan (a)

2 πn < a < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < a < 2 π + 2 πn

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