What is the general solution for 2.1sin(θ)=1.33sin(θ+90) ?

The general solution for 2.1sin(θ)=1.33sin(θ+90) is θ=0.56456…+180n

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2.1sin(θ)=1.33sin(θ+90)

Solution

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

Solution

θ = 0.56456 \dots + 18 0^{\circ} n

Radians

θ = 0.56456 \dots + πn

Solution steps

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

Rewrite using trig identities

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

Rewrite using trig identities

sin (θ + 9 0^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (θ) cos (9 0^{\circ}) + cos (θ) sin (9 0^{\circ})

Simplify

sin (θ) cos (9 0^{\circ}) + cos (θ) sin (9 0^{\circ}) : cos (θ)

sin (θ) cos (9 0^{\circ}) + cos (θ) sin (9 0^{\circ})

sin (θ) cos (9 0^{\circ}) = 0

sin (θ) cos (9 0^{\circ})

Simplify

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0 \cdot sin (θ)

Apply rule

0 \cdot a = 0

= 0

cos (θ) sin (9 0^{\circ}) = cos (θ)

cos (θ) sin (9 0^{\circ})

Simplify

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1 \cdot cos (θ)

Multiply:

cos (θ) \cdot 1 = cos (θ)

= cos (θ)

= 0 + cos (θ)

0 + cos (θ) = cos (θ)

= cos (θ)

= cos (θ)

2.1 sin (θ) = 1.33 cos (θ)

2.1 sin (θ) = 1.33 cos (θ)

Subtract

1.33 cos (θ)

from both sides

2.1 sin (θ) - 1.33 cos (θ) = 0

Rewrite using trig identities

2.1 sin (θ) - 1.33 cos (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{2.1 sin ( θ ) - 1.33 cos ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

\frac{2.1 sin ( θ )}{cos ( θ )} - 1.33 = 0

Use the basic trigonometric identity:

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

2.1 tan (θ) - 1.33 = 0

2.1 tan (θ) - 1.33 = 0

Multiply both sides by

100

2.1 tan (θ) - 1.33 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

2.1 tan (θ) \cdot 100 - 1.33 \cdot 100 = 0 \cdot 100

Refine

210 tan (θ) - 133 = 0

210 tan (θ) - 133 = 0

Move

133

to the right side

210 tan (θ) - 133 = 0

Add

133

to both sides

210 tan (θ) - 133 + 133 = 0 + 133

Simplify

210 tan (θ) = 133

210 tan (θ) = 133

Divide both sides by

210

210 tan (θ) = 133

Divide both sides by

210

\frac{210 tan ( θ )}{210} = \frac{133}{210}

Simplify

tan (θ) = \frac{19}{30}

tan (θ) = \frac{19}{30}

Apply trig inverse properties

tan (θ) = \frac{19}{30}

General solutions for

tan (θ) = \frac{19}{30}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

θ = arctan (\frac{19}{30}) + 18 0^{\circ} n

θ = arctan (\frac{19}{30}) + 18 0^{\circ} n

Show solutions in decimal form

θ = 0.56456 \dots + 18 0^{\circ} n

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tan (6 t) = 2

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Frequently Asked Questions (FAQ)

What is the general solution for 2.1sin(θ)=1.33sin(θ+90) ?
The general solution for 2.1sin(θ)=1.33sin(θ+90) is θ=0.56456…+180n