Is sin(θ)=-cos(θ+pi/2) ?

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

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prove sin(θ)=-cos(θ+pi/2)

prove

sin (θ) = - cos (θ + \frac{π}{2})

True

Solution steps

sin (θ) = - cos (θ + \frac{π}{2})

Manipulating right side

- cos (θ + \frac{π}{2})

Rewrite using trig identities

cos (θ + \frac{π}{2})

Use the Angle Sum identity:

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

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

Simplify

cos (θ) cos (\frac{π}{2}) - sin (θ) sin (\frac{π}{2}) : - sin (θ)

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

cos (θ) cos (\frac{π}{2}) = 0

cos (θ) cos (\frac{π}{2})

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 (θ)

Apply rule

0 \cdot a = 0

= 0

sin (θ) sin (\frac{π}{2}) = sin (θ)

sin (θ) sin (\frac{π}{2})

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 (θ)

Multiply:

sin (θ) \cdot 1 = sin (θ)

= sin (θ)

= 0 - sin (θ)

0 - sin (θ) = - sin (θ)

= - sin (θ)

= - sin (θ)

= - (- sin (θ))

Apply rule

- (- a) = a

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( θ ) + cos ( θ )}{sin ( θ )} = 1 + cot (θ)

cos^{2} (x) (sec (x) + 1)^{2} = (cos (x) + 1)^{2}

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

\frac{cos ( u )}{1 + tan ( u )} - \frac{sin ( u )}{1 + cot ( u )} = cos (u) - sin (u)

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