Is cos(pi/7)=sin(pi/2-pi/7) ?

The answer to whether cos(pi/7)=sin(pi/2-pi/7) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos(pi/7)=sin(pi/2-pi/7)

prove

cos (\frac{π}{7}) = sin (\frac{π}{2} - \frac{π}{7})

True

Solution steps

cos (\frac{π}{7}) = sin (\frac{π}{2} - \frac{π}{7})

Manipulating right side

sin (\frac{π}{2} - \frac{π}{7})

Rewrite using trig identities

sin (\frac{π}{2} - \frac{π}{7})

Use the Angle Difference identity:

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

= sin (\frac{π}{2}) cos (\frac{π}{7}) - cos (\frac{π}{2}) sin (\frac{π}{7})

sin (\frac{π}{2}) cos (\frac{π}{7}) - cos (\frac{π}{2}) sin (\frac{π}{7}) = cos (\frac{π}{7})

sin (\frac{π}{2}) cos (\frac{π}{7}) - cos (\frac{π}{2}) sin (\frac{π}{7})

sin (\frac{π}{2}) cos (\frac{π}{7}) = cos (\frac{π}{7})

sin (\frac{π}{2}) cos (\frac{π}{7})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (\frac{π}{7})

Multiply:

1 \cdot cos (\frac{π}{7}) = cos (\frac{π}{7})

= cos (\frac{π}{7})

cos (\frac{π}{2}) sin (\frac{π}{7}) = 0

cos (\frac{π}{2}) sin (\frac{π}{7})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (\frac{π}{7})

Apply rule

0 \cdot a = 0

= 0

= cos (\frac{π}{7}) - 0

cos (\frac{π}{7}) - 0 = cos (\frac{π}{7})

= cos (\frac{π}{7})

= cos (\frac{π}{7})

= cos (\frac{π}{7})

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{π}{4} + x) = 2 (cos (x) + sin (x))

sec (x) csc (x) = cot (x) + tan (x)

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

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

1 + sin (4 a) + sin (2 a) = - 2 cos (a)

Is cos(pi/7)=sin(pi/2-pi/7) ?
The answer to whether cos(pi/7)=sin(pi/2-pi/7) is True