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

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

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

prove

cos (\frac{π}{2} - a) = sin (a)

True

Solution steps

cos (\frac{π}{2} - a) = sin (a)

Manipulating left side

cos (\frac{π}{2} - a)

Rewrite using trig identities

cos (\frac{π}{2} - a)

Use the Angle Difference identity:

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

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

Simplify

cos (\frac{π}{2}) cos (a) + sin (\frac{π}{2}) sin (a) : sin (a)

cos (\frac{π}{2}) cos (a) + sin (\frac{π}{2}) sin (a)

cos (\frac{π}{2}) cos (a) = 0

cos (\frac{π}{2}) cos (a)

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 cos (a)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (a) = sin (a)

sin (\frac{π}{2}) sin (a)

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 sin (a)

Multiply:

1 \cdot sin (a) = sin (a)

= sin (a)

= 0 + sin (a)

0 + sin (a) = sin (a)

= sin (a)

= sin (a)

= sin (a)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x + y )}{sin ( x ) cos ( y )} = 1 + cot (x) tan (y)

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

sin^{2} (2 x) = \frac{1}{2} - \frac{1}{2} cos (4 x)

sin (\frac{a}{2}) = \frac{1}{2} sin (a)

tan^{2} (α) = \frac{sin ^{2} ( α )}{1 - sin ^{2} ( α )}