Is cos(θ+30)-sin(θ+60)=-sin(θ) ?

The answer to whether cos(θ+30)-sin(θ+60)=-sin(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos(θ+30)-sin(θ+60)=-sin(θ)

Solution

prove

cos (θ + 3 0^{\circ}) - sin (θ + 6 0^{\circ}) = - sin (θ)

Solution

True

Solution steps

cos (θ + 3 0^{\circ}) - sin (θ + 6 0^{\circ}) = - sin (θ)

Manipulating left side

cos (θ + 3 0^{\circ}) - sin (θ + 6 0^{\circ})

Rewrite using trig identities

sin (θ + 6 0^{\circ})

Use the Angle Sum identity:

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

= sin (θ) cos (6 0^{\circ}) + cos (θ) sin (6 0^{\circ})

Simplify

sin (θ) cos (6 0^{\circ}) + cos (θ) sin (6 0^{\circ}) : \frac{1}{2} sin (θ) + \frac{3}{2} cos (θ)

sin (θ) cos (6 0^{\circ}) + cos (θ) sin (6 0^{\circ})

Simplify

cos (6 0^{\circ}) : \frac{1}{2}

cos (6 0^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

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}

= \frac{1}{2}

= \frac{1}{2} sin (θ) + sin (6 0^{\circ}) cos (θ)

Simplify

sin (6 0^{\circ}) : \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

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}

= \frac{3}{2}

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

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

= cos (θ + 3 0^{\circ}) - (\frac{1}{2} sin (θ) + \frac{3}{2} cos (θ))

Rewrite using trig identities

cos (θ + 3 0^{\circ})

Use the Angle Sum identity:

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

= cos (θ) cos (3 0^{\circ}) - sin (θ) sin (3 0^{\circ})

Simplify

cos (θ) cos (3 0^{\circ}) - sin (θ) sin (3 0^{\circ}) : \frac{3}{2} cos (θ) - \frac{1}{2} sin (θ)

cos (θ) cos (3 0^{\circ}) - sin (θ) sin (3 0^{\circ})

Simplify

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

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}

= \frac{3}{2}

= \frac{3}{2} cos (θ) - sin (3 0^{\circ}) sin (θ)

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

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}

= \frac{1}{2}

= \frac{3}{2} cos (θ) - \frac{1}{2} sin (θ)

= \frac{3}{2} cos (θ) - \frac{1}{2} sin (θ)

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

Simplify

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

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

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

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

Distribute parentheses

= - (\frac{1}{2} sin (θ)) - (\frac{3}{2} cos (θ))

Apply minus-plus rules

+ (- a) = - a

= - \frac{1}{2} sin (θ) - \frac{3}{2} cos (θ)

= \frac{3}{2} cos (θ) - \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ) - \frac{3}{2} cos (θ)

Simplify

\frac{3}{2} cos (θ) - \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ) - \frac{3}{2} cos (θ) : - sin (θ)

\frac{3}{2} cos (θ) - \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ) - \frac{3}{2} cos (θ)

Group like terms

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

Add similar elements:

\frac{3}{2} cos (θ) - \frac{3}{2} cos (θ) = 0

\frac{3}{2} cos (θ) - \frac{3}{2} cos (θ)

Factor out common term

cos (θ)

= cos (θ) (\frac{3}{2} - \frac{3}{2})

\frac{3}{2} - \frac{3}{2} = 0

\frac{3}{2} - \frac{3}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 3}{2}

Factor

3 - 3 : 0

3 - 3

Factor out common term

3

= 3 (1 - 1)

Refine

= 0

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

= 0

= - \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ)

Add similar elements:

- \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ) = - sin (θ)

- \frac{1}{2} sin (θ) - \frac{1}{2} sin (θ)

Factor out common term

sin (θ)

= sin (θ) (- \frac{1}{2} - \frac{1}{2})

- \frac{1}{2} - \frac{1}{2} = - 1

- \frac{1}{2} - \frac{1}{2}

Apply rule

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

= \frac{- 1 - 1}{2}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

= - sin (θ)

= - sin (θ)

= - sin (θ)

= - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (tan^2(x))/(sec(x))=sin(tan(x))prove

\frac{tan ^{2} ( x )}{sec ( x )} = sin (tan (x))

prove sin^2(0)=2-2cos(0)prove

sin^{2} (0) = 2 - 2 cos (0)

prove cos((3pi)/8)=(sqrt(2-\sqrt{2)})/2 prove

cos (\frac{3 π}{8}) = \frac{2 - 2}{2}

prove tan(a)*cot(a)=sin^2(a)+cos^2(a)prove

tan (a) \cdot cot (a) = sin^{2} (a) + cos^{2} (a)

prove tan(x)+(cos(x))/(1-sin(x))=sec(x)prove

tan (x) + \frac{cos ( x )}{1 - sin ( x )} = sec (x)

Frequently Asked Questions (FAQ)

Is cos(θ+30)-sin(θ+60)=-sin(θ) ?
The answer to whether cos(θ+30)-sin(θ+60)=-sin(θ) is True