What is the general solution for solvefor A,cos(pi/3-A)=2sin(A) ?

The general solution for solvefor A,cos(pi/3-A)=2sin(A) is A=0.41528…+pin

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solvefor A,cos(pi/3-A)=2sin(A)

Solution

solve for

A, cos (\frac{π}{3} - A) = 2 sin (A)

Solution

A = 0.41528 \dots + πn

Degrees

A = 23.79397 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (\frac{π}{3} - A) = 2 sin (A)

Rewrite using trig identities

cos (\frac{π}{3} - A) = 2 sin (A)

Rewrite using trig identities

cos (\frac{π}{3} - A)

Use the Angle Difference identity:

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

= cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A)

Simplify

cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A) : \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A)

Simplify

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

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}

= \frac{1}{2}

= \frac{1}{2} cos (A) + sin (\frac{π}{3}) sin (A)

Simplify

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{2}

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}

= \frac{3}{2}

= \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

= \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

\frac{1}{2} cos (A) + \frac{3}{2} sin (A) = 2 sin (A)

\frac{1}{2} cos (A) + \frac{3}{2} sin (A) = 2 sin (A)

Subtract

2 sin (A)

from both sides

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A) = 0

Simplify

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A) : \frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2}

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A)

\frac{1}{2} cos (A) = \frac{cos ( A )}{2}

\frac{1}{2} cos (A)

Multiply fractions:

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

= \frac{1 \cdot cos ( A )}{2}

Multiply:

1 \cdot cos (A) = cos (A)

= \frac{cos ( A )}{2}

\frac{3 - 4}{2} sin (A) = \frac{( 3 - 4 ) sin ( A )}{2}

\frac{3 - 4}{2} sin (A)

Multiply fractions:

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

= \frac{( 3 - 4 ) sin ( A )}{2}

= \frac{cos ( A )}{2} + \frac{( 3 - 4 ) sin ( A )}{2}

Apply rule

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

= \frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2}

\frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2} = 0

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

cos (A) + (3 - 4) sin (A) = 0

Rewrite using trig identities

cos (A) + (3 - 4) sin (A) = 0

Divide both sides by

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

\frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{cos ( A )} = \frac{0}{cos ( A )}

Simplify

1 + \frac{3 sin ( A )}{cos ( A )} - \frac{4 sin ( A )}{cos ( A )} = 0

Use the basic trigonometric identity:

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

1 + (- 4 + 3) tan (A) = 0

1 + (- 4 + 3) tan (A) = 0

Move

1

to the right side

1 + (- 4 + 3) tan (A) = 0

Subtract

1

from both sides

1 + (- 4 + 3) tan (A) - 1 = 0 - 1

Simplify

(- 4 + 3) tan (A) = - 1

(- 4 + 3) tan (A) = - 1

Divide both sides by

- 4 + 3

(- 4 + 3) tan (A) = - 1

Divide both sides by

- 4 + 3

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} = \frac{- 1}{- 4 + 3}

Simplify

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} = \frac{- 1}{- 4 + 3}

Simplify

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} : tan (A)

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3}

Cancel the common factor:

- 4 + 3

= tan (A)

Simplify

\frac{- 1}{- 4 + 3} : \frac{4 + 3}{13}

\frac{- 1}{- 4 + 3}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

- 4 + 3 = - (4 - 3)

= \frac{1}{4 - 3}

Rationalize

\frac{1}{4 - 3} : \frac{4 + 3}{13}

\frac{1}{4 - 3}

Multiply by the conjugate

\frac{4 + 3}{4 + 3}

= \frac{1 \cdot ( 4 + 3 )}{( 4 - 3 ) ( 4 + 3 )}

1 \cdot (4 + 3) = 4 + 3

(4 - 3) (4 + 3) = 13

(4 - 3) (4 + 3)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = 4, b = 3

= 4^{2} - (3)^{2}

Simplify

4^{2} - (3)^{2} : 13

4^{2} - (3)^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 16 - 3

Subtract the numbers:

16 - 3 = 13

= 13

= 13

= \frac{4 + 3}{13}

= \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

Apply trig inverse properties

tan (A) = \frac{4 + 3}{13}

General solutions for

tan (A) = \frac{4 + 3}{13}

tan (x) = a \Rightarrow x = arctan (a) + πn

A = arctan (\frac{4 + 3}{13}) + πn

A = arctan (\frac{4 + 3}{13}) + πn

Show solutions in decimal form

A = 0.41528 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor A,cos(pi/3-A)=2sin(A) ?
The general solution for solvefor A,cos(pi/3-A)=2sin(A) is A=0.41528…+pin