Is cos(pi/2+y)=-sin(y) ?

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

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prove cos(pi/2+y)=-sin(y)

prove

cos (\frac{π}{2} + y) = - sin (y)

True

Solution steps

cos (\frac{π}{2} + y) = - sin (y)

Manipulating left side

cos (\frac{π}{2} + y)

Rewrite using trig identities

cos (\frac{π}{2} + y)

Use the Angle Sum identity:

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

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

Simplify

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

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

cos (\frac{π}{2}) cos (y) = 0

cos (\frac{π}{2}) cos (y)

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 (y)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (y) = sin (y)

sin (\frac{π}{2}) sin (y)

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 (y)

Multiply:

1 \cdot sin (y) = sin (y)

= sin (y)

= 0 - sin (y)

0 - sin (y) = - sin (y)

= - sin (y)

= - sin (y)

= - sin (y)

We showed that the two sides could take the same form

\Rightarrow True

- tan^{2} (x) + sec^{2} (x) = 1

tan (2 θ) = \frac{2 cot ( θ )}{cot ^{2} ( θ ) - 1}

csc (x) - tan (\frac{x}{2}) = cot (x)

cos (3 x) - cos (x) = 4 cos^{3} (x) - 4 cos (x)

cot^{2} (θ) - cos^{2} (θ) = cos^{2} (θ) cot^{2} (θ)