Is cos((3x)/5)=-sin((3x)/5-pi/2) ?

The answer to whether cos((3x)/5)=-sin((3x)/5-pi/2) is True

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prove cos((3x)/5)=-sin((3x)/5-pi/2)

prove

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

True

Solution steps

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

Manipulating right side

- sin (\frac{3 x}{5} - \frac{π}{2})

Rewrite using trig identities

sin (\frac{3 x}{5} - \frac{π}{2})

Use the Angle Difference identity:

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

= sin (\frac{3 x}{5}) cos (\frac{π}{2}) - cos (\frac{3 x}{5}) sin (\frac{π}{2})

Simplify

sin (\frac{3 x}{5}) cos (\frac{π}{2}) - cos (\frac{3 x}{5}) sin (\frac{π}{2}) : - cos (\frac{3 x}{5})

sin (\frac{3 x}{5}) cos (\frac{π}{2}) - cos (\frac{3 x}{5}) sin (\frac{π}{2})

sin (\frac{3 x}{5}) cos (\frac{π}{2}) = 0

sin (\frac{3 x}{5}) 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 sin (\frac{3 x}{5})

Apply rule

0 \cdot a = 0

= 0

cos (\frac{3 x}{5}) sin (\frac{π}{2}) = cos (\frac{3 x}{5})

cos (\frac{3 x}{5}) 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 cos (\frac{3 x}{5})

Multiply:

cos (\frac{3 x}{5}) \cdot 1 = cos (\frac{3 x}{5})

= cos (\frac{3 x}{5})

= 0 - cos (\frac{3 x}{5})

0 - cos (\frac{3 x}{5}) = - cos (\frac{3 x}{5})

= - cos (\frac{3 x}{5})

= - cos (\frac{3 x}{5})

= - (- cos (\frac{3 x}{5}))

Apply rule

- (- a) = a

= cos (\frac{3 x}{5})

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

\frac{sin ^{2} ( θ )}{1 + cos ( θ )} = 1 - cos (θ)

sin^{2} (x) = 3 cos^{2} (x)

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

Is cos((3x)/5)=-sin((3x)/5-pi/2) ?
The answer to whether cos((3x)/5)=-sin((3x)/5-pi/2) is True