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

The answer to whether sin(x-pi/2)=cos(x) is False

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

prove

sin (x - \frac{π}{2}) = cos (x)

False

Solution steps

sin (x - \frac{π}{2}) = cos (x)

Show that the two sides are not equal

For

sin (x - \frac{π}{2}) = cos (x)

plug in

x = 0

sin (0 - \frac{π}{2}) = - 1

sin (0 - \frac{π}{2})

Simplify

= sin (- \frac{π}{2})

Use the following property:

sin (- x) = - sin (x)

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

= - sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

= - 1

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (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}

= 1

= 1

The two sides are not equal

\Rightarrow False

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

2 csc (2 x) = sec^{2} (x) cot (x)

cos^{2} (x) - cos^{2} (x) = 0

2 (sec (x) - tan (x)) (csc (x) + 1) = 2 cot (x)

sec (x) sin (x) cot (x) = 1