Is sin(67)=sin(42)cos(25)+cos(42)sin(25) ?

The answer to whether sin(67)=sin(42)cos(25)+cos(42)sin(25) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(67)=sin(42)cos(25)+cos(42)sin(25)

Solution

prove

sin (6 7^{\circ}) = sin (4 2^{\circ}) cos (2 5^{\circ}) + cos (4 2^{\circ}) sin (2 5^{\circ})

Solution

True

Solution steps

sin (6 7^{\circ}) = sin (4 2^{\circ}) cos (2 5^{\circ}) + cos (4 2^{\circ}) sin (2 5^{\circ})

Manipulating right side

sin (4 2^{\circ}) cos (2 5^{\circ}) + cos (4 2^{\circ}) sin (2 5^{\circ})

Rewrite using trig identities

sin (4 2^{\circ}) cos (2 5^{\circ}) + cos (4 2^{\circ}) sin (2 5^{\circ})

Use the Angle Sum identity:

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

= sin (4 2^{\circ} + 2 5^{\circ})

Join

4 2^{\circ} + 2 5^{\circ} : 6 7^{\circ}

4 2^{\circ} + 2 5^{\circ}

Least Common Multiplier of

30, 36 : 180

30, 36

Least Common Multiplier (LCM)

Prime factorization of

30 : 2 \cdot 3 \cdot 5

30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 5

Prime factorization of

36 : 2 \cdot 2 \cdot 3 \cdot 3

36

36

divides by

236 = 18 \cdot 2

= 2 \cdot 18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 3

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

30

36

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5

Multiply the numbers:

2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 = 180

= 180

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

180

For

4 2^{\circ} :

multiply the denominator and numerator by

6

4 2^{\circ} = \frac{126 0 ^{\circ} 6}{30 \cdot 6} = 4 2^{\circ}

For

2 5^{\circ} :

multiply the denominator and numerator by

5

2 5^{\circ} = \frac{90 0 ^{\circ} 5}{36 \cdot 5} = 2 5^{\circ}

= 4 2^{\circ} + 2 5^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{756 0 ^{\circ} + 450 0 ^{\circ}}{180}

Add similar elements:

756 0^{\circ} + 450 0^{\circ} = 1206 0^{\circ}

= 6 7^{\circ}

= sin (6 7^{\circ})

= sin (6 7^{\circ})

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove cos(90+30)=cos(120)prove

cos (9 0^{\circ} + 3 0^{\circ}) = cos (12 0^{\circ})

prove sin(θ)cos(θ)=sin(θ)prove

sin (θ) cos (θ) = sin (θ)

prove (1-cos(x))-cos(y)=sin(x+y)prove

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

prove 1-sin(x)=cos^2(x)prove

1 - sin (x) = cos^{2} (x)

prove sin^2(θ)(1+cot^2(θ))-1=0 prove

sin^{2} (θ) (1 + cot^{2} (θ)) - 1 = 0

Frequently Asked Questions (FAQ)

Is sin(67)=sin(42)cos(25)+cos(42)sin(25) ?
The answer to whether sin(67)=sin(42)cos(25)+cos(42)sin(25) is True