Is csc(pi/2-x)cos(x)=1 ?

The answer to whether csc(pi/2-x)cos(x)=1 is True

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prove csc(pi/2-x)cos(x)=1

prove

csc (\frac{π}{2} - x) cos (x) = 1

True

Solution steps

csc (\frac{π}{2} - x) cos (x) = 1

Manipulating left side

csc (\frac{π}{2} - x) cos (x)

Rewrite using trig identities

csc (\frac{π}{2} - x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( \frac{π}{2} - x )}

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

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 (x)

Multiply:

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

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

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 (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

= \frac{1}{cos ( x )}

= \frac{1}{cos ( x )}

= \frac{1}{cos ( x )} cos (x)

\frac{1}{cos ( x )} cos (x) = 1

\frac{1}{cos ( x )} cos (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

cos (x)

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

1 - 2 cos^{2} (4) = 2 sin^{2} (4) - 1

\frac{tan ( θ )}{cot ( θ ) + tan ( θ )} = sin^{2} (θ)

2 sin (x) + (2 + 2) = - 2 csc (x)

sin (8 a) = 4 sin (2 a) cos (2 a) cos (4 a)

sin^{2} (2 x) = \frac{1}{2 ( 1 - cos ^{4} ( x ) )}