Is sin(pi/6+b)+cos(pi/3+b)=cos(b) ?

The answer to whether sin(pi/6+b)+cos(pi/3+b)=cos(b) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(pi/6+b)+cos(pi/3+b)=cos(b)

Solution

prove

sin (\frac{π}{6} + b) + cos (\frac{π}{3} + b) = cos (b)

Solution

True

Solution steps

sin (\frac{π}{6} + b) + cos (\frac{π}{3} + b) = cos (b)

Manipulating left side

sin (\frac{π}{6} + b) + cos (\frac{π}{3} + b)

Rewrite using trig identities

cos (\frac{π}{3} + b)

Use the Angle Sum identity:

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

= cos (\frac{π}{3}) cos (b) - sin (\frac{π}{3}) sin (b)

Simplify

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

cos (\frac{π}{3}) cos (b) - sin (\frac{π}{3}) sin (b)

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 (b) - sin (\frac{π}{3}) sin (b)

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 (b) - \frac{3}{2} sin (b)

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

= sin (\frac{π}{6} + b) + \frac{1}{2} cos (b) - \frac{3}{2} sin (b)

Rewrite using trig identities

sin (\frac{π}{6} + b)

Use the Angle Sum identity:

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

= sin (\frac{π}{6}) cos (b) + cos (\frac{π}{6}) sin (b)

Simplify

sin (\frac{π}{6}) cos (b) + cos (\frac{π}{6}) sin (b) : \frac{1}{2} cos (b) + \frac{3}{2} sin (b)

sin (\frac{π}{6}) cos (b) + cos (\frac{π}{6}) sin (b)

Simplify

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{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{1}{2}

= \frac{1}{2} cos (b) + cos (\frac{π}{6}) sin (b)

Simplify

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{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{3}{2}

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

\frac{1}{2} cos (b) + \frac{1}{2} cos (b) = cos (b)

\frac{1}{2} cos (b) + \frac{1}{2} cos (b)

Factor out common term

cos (b)

= cos (b) (\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}

Refine

= 1

= cos (b)

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

Add similar elements:

\frac{3}{2} sin (b) - \frac{3}{2} sin (b) = 0

\frac{3}{2} sin (b) - \frac{3}{2} sin (b)

Factor out common term

sin (b)

= sin (b) (\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

= cos (b)

= cos (b)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

identity cos^2(x)-sin^2(x)identity

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

identity 1-cos^2(x)identity

1 - cos^{2} (x)

identity 1-sin^2(x)identity

1 - sin^{2} (x)

identity sin(2θ)identity

sin (2 θ)

identity tan^2(x)identity

tan^{2} (x)

Frequently Asked Questions (FAQ)

Is sin(pi/6+b)+cos(pi/3+b)=cos(b) ?
The answer to whether sin(pi/6+b)+cos(pi/3+b)=cos(b) is True