Is cos(2x-90)=2sin(x)cos(x) ?

The answer to whether cos(2x-90)=2sin(x)cos(x) is False

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prove cos(2x-90)=2sin(x)cos(x)

prove

cos (2 x - 90) = 2 sin (x) cos (x)

False

Solution steps

cos (2 x - 90) = 2 sin (x) cos (x)

Show that the two sides are not equal

For

cos (2 x - 90) = 2 sin (x) cos (x)

plug in

x = 0

cos (2 \cdot 0 - 90) = - 0.44807 \dots

cos (2 \cdot 0 - 90)

Refine to a decimal form

= - 0.44807 \dots

2 sin (0) cos (0) = 0

2 sin (0) cos (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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}

= 0

Use the following trivial identity:

cos (0) = 1

cos (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}

= 1

= 2 \cdot 0 \cdot 1

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e sin^{4} (x) - 2 sin^{2} (x) = cos^{4} (x) - 1

p ro v e 1 + sec^{2} (x) sin (x) = sec^{2} (x)

p ro v e \frac{cot ( x )}{csc ( x ) - sin ( x )} = sec (x)

p ro v e sin (x) = 5 cos (x)

p ro v e \frac{sin ( 3 x ) + sin ( 5 x )}{cos ( 3 x ) - cos ( 5 x )} = cot (x)

Is cos(2x-90)=2sin(x)cos(x) ?
The answer to whether cos(2x-90)=2sin(x)cos(x) is False