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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

sin (- x) cos (\frac{π}{2} - x)

Use the negative angle identity:

sin (- x) = - sin (x)

= cos (\frac{π}{2} - x) (- sin (x))

Rewrite using trig identities

cos (\frac{π}{2} - x)

Use the Angle Difference identity:

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

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

Simplify

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

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

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (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 cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (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 sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

= sin (x)

= sin (x) (- sin (x))

Simplify

sin (x) (- sin (x)) : - sin^{2} (x)

sin (x) (- sin (x))

Remove parentheses:

(- a) = - a

= - sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= - sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= - sin^{2} (x)

= - sin^{2} (x)

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

1 - 2 sin^{2} (x) - 1

Group like terms

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

1 - 1 = 0

= - 2 sin^{2} (x)

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

Divide the numbers:

\frac{2}{2} = 1

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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