Is cos(pi/2)=sin(pi/2)-1 ?

The answer to whether cos(pi/2)=sin(pi/2)-1 is True

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prove cos(pi/2)=sin(pi/2)-1

prove

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

True

Solution steps

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

Manipulating left side

cos (\frac{π}{2})

Simplify

= 0

Manipulating right side

sin (\frac{π}{2}) - 1

Simplify

- 1 + sin (\frac{π}{2}) : 0

- 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

Simplify

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) = csc^{2} (x) \cdot tan^{2} (x) - 1

\frac{1 + tan ( θ )}{sin ( θ )} - sec (θ) = csc (θ)

cot (a) = tan (\frac{π}{2} - a)

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

\frac{2}{sec ^{2} ( x )} + \frac{2}{csc ^{2} ( x )} = 2