What is the value of cos(300)cos(180)-sin(180)sin(300) ?

The value of cos(300)cos(180)-sin(180)sin(300) is -1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(300)cos(180)-sin(180)sin(300)

cos (30 0^{\circ}) cos (18 0^{\circ}) - sin (18 0^{\circ}) sin (30 0^{\circ})

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

cos (30 0^{\circ}) cos (18 0^{\circ}) - sin (18 0^{\circ}) sin (30 0^{\circ})

Rewrite as

= cos (30 0^{\circ}) cos (18 0^{\circ}) - sin (30 0^{\circ}) sin (18 0^{\circ})

Rewrite using trig identities:

cos (12 0^{\circ})

cos (30 0^{\circ}) cos (18 0^{\circ}) - sin (30 0^{\circ}) sin (18 0^{\circ})

Use the Angle Sum identity:

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

= cos (30 0^{\circ} + 18 0^{\circ})

Simplify

= cos (48 0^{\circ})

cos (48 0^{\circ}) = cos (12 0^{\circ})

cos (48 0^{\circ})

Rewrite

48 0^{\circ}

36 0^{\circ} + 12 0^{\circ}

= cos (36 0^{\circ} + 12 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ}) = cos (x)

cos (36 0^{\circ} + 12 0^{\circ}) = cos (12 0^{\circ})

= cos (12 0^{\circ})

= cos (12 0^{\circ})

= cos (12 0^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

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}

4 cos (6 5^{\circ})

\frac{7 sin ( - 0.1 )}{- 0.1}

\frac{( 6 ) ^{2} sin ( 2 ) ( 3 0 ^{\circ} )}{9.8}

12 \cdot tan (3 0^{\circ})

arcsin (0.952)

What is the value of cos(300)cos(180)-sin(180)sin(300) ?
The value of cos(300)cos(180)-sin(180)sin(300) is -1/2