Is sin((5pi)/2+x)=cos(x) ?

The answer to whether sin((5pi)/2+x)=cos(x) is True

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prove sin((5pi)/2+x)=cos(x)

prove

sin (\frac{5 π}{2} + x) = cos (x)

True

Solution steps

sin (\frac{5 π}{2} + x) = cos (x)

Manipulating left side

sin (\frac{5 π}{2} + x)

Rewrite using trig identities

sin (\frac{5 π}{2} + x)

Use the Angle Sum identity:

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

= sin (\frac{5 π}{2}) cos (x) + cos (\frac{5 π}{2}) sin (x)

Simplify

sin (\frac{5 π}{2}) cos (x) + cos (\frac{5 π}{2}) sin (x) : cos (x)

sin (\frac{5 π}{2}) cos (x) + cos (\frac{5 π}{2}) sin (x)

sin (\frac{5 π}{2}) cos (x) = cos (x)

sin (\frac{5 π}{2}) cos (x)

sin (\frac{5 π}{2}) = 1

sin (\frac{5 π}{2})

sin (\frac{5 π}{2}) = sin (\frac{π}{2})

sin (\frac{5 π}{2})

Rewrite

\frac{5 π}{2}

2 π + \frac{π}{2}

= sin (2 π + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + \frac{π}{2}) = 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

= 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

cos (\frac{5 π}{2}) sin (x) = 0

cos (\frac{5 π}{2}) sin (x)

cos (\frac{5 π}{2}) = 0

cos (\frac{5 π}{2})

cos (\frac{5 π}{2}) = cos (\frac{π}{2})

cos (\frac{5 π}{2})

Rewrite

\frac{5 π}{2}

2 π + \frac{π}{2}

= cos (2 π + \frac{π}{2})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

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

= cos (\frac{π}{2})

= cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{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}

= 0

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) + 0

cos (x) + 0 = cos (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{2}{cos ^{2} ( x ) - tan ^{2} ( x )} = 2 + tan^{2} (x)

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

tan^{2} (\frac{π}{4}) = tan (\frac{π}{4})

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

sec (A) - tan (A) = \frac{cos ( A )}{1 + sin ( A )}