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

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

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

prove

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

True

Solution steps

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

Manipulating right side

2 cos (x) cos (\frac{π}{2} - 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)

= 2 cos (x) sin (x)

Rewrite using trig identities

2 cos (x) sin (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= sin (2 x)

= sin (2 x)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) = \frac{1 + tan ^{2} ( x )}{1 + cot ^{2} ( x )}

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

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

tan (x) = cot (x) - 2 cot (2 x)

tan (3 x) (tan (x) - 1) = 0

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