What is the general solution for (sin(54))/7 =(sin(x))/(10) ?

The general solution for (sin(54))/7 =(sin(x))/(10) is No Solution for x\in\mathbb{R}

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Popular Trigonometry >

(sin(54))/7 =(sin(x))/(10)

Solution

\frac{sin ( 5 4 ^{\circ} )}{7} = \frac{sin ( x )}{10}

Solution

No Solution for x \in R

Solution steps

\frac{sin ( 5 4 ^{\circ} )}{7} = \frac{sin ( x )}{10}

sin (5 4^{\circ}) = \frac{5 + 1}{4}

sin (5 4^{\circ})

Rewrite using trig identities:

cos (3 6^{\circ})

sin (5 4^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

= cos (9 0^{\circ} - 5 4^{\circ})

Simplify:

9 0^{\circ} - 5 4^{\circ} = 3 6^{\circ}

9 0^{\circ} - 5 4^{\circ}

Least Common Multiplier of

2, 10 : 10

2, 10

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

10 : 2 \cdot 5

10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 5

Multiply each factor the greatest number of times it occurs in either

2

10

= 2 \cdot 5

Multiply the numbers:

2 \cdot 5 = 10

= 10

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

10

For

9 0^{\circ} :

multiply the denominator and numerator by

5

9 0^{\circ} = \frac{18 0 ^{\circ} 5}{2 \cdot 5} = 9 0^{\circ}

= 9 0^{\circ} - 5 4^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 5 - 54 0 ^{\circ}}{10}

Add similar elements:

90 0^{\circ} - 54 0^{\circ} = 36 0^{\circ}

= 3 6^{\circ}

Cancel the common factor:

2

= 3 6^{\circ}

= cos (3 6^{\circ})

= cos (3 6^{\circ})

Rewrite using trig identities:

\frac{5 + 1}{4}

cos (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{5 + 1}{4}

\frac{\frac{5 + 1}{4}}{7} = \frac{sin ( x )}{10}

Switch sides

\frac{sin ( x )}{10} = \frac{\frac{5 + 1}{4}}{7}

Multiply both sides by

10

\frac{sin ( x )}{10} = \frac{\frac{5 + 1}{4}}{7}

Multiply both sides by

10

\frac{sin ( x )}{10} \cdot 10 = \frac{\frac{5 + 1}{4}}{7} \cdot 10

Simplify

\frac{sin ( x )}{10} \cdot 10 = \frac{\frac{5 + 1}{4}}{7} \cdot 10

Simplify

\frac{sin ( x )}{10} \cdot 10 : sin (x)

\frac{sin ( x )}{10} \cdot 10

Multiply fractions:

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

= \frac{sin ( x ) \cdot 10}{10}

Cancel the common factor:

10

= sin (x)

Simplify

\frac{\frac{5 + 1}{4}}{7} \cdot 10 : \frac{5 ( 1 + 5 )}{14}

\frac{\frac{5 + 1}{4}}{7} \cdot 10

\frac{\frac{5 + 1}{4}}{7} = \frac{5 + 1}{28}

\frac{\frac{5 + 1}{4}}{7}

Apply the fraction rule:

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

= \frac{5 + 1}{4 \cdot 7}

Multiply the numbers:

4 \cdot 7 = 28

= \frac{5 + 1}{28}

= 10 \cdot \frac{1 + 5}{28}

Multiply fractions:

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

= \frac{( 5 + 1 ) \cdot 10}{28}

Cancel the common factor:

2

= \frac{5 ( 1 + 5 )}{14}

sin (x) = \frac{5 ( 1 + 5 )}{14}

sin (x) = \frac{5 ( 1 + 5 )}{14}

sin (x) = \frac{5 ( 1 + 5 )}{14}

- 1 \leq sin (x) \leq 1

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(54))/7 =(sin(x))/(10) ?
The general solution for (sin(54))/7 =(sin(x))/(10) is No Solution for x\in\mathbb{R}