Is 1+cos(2x)=3sin(x) ?

The answer to whether 1+cos(2x)=3sin(x) is False

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prove 1+cos(2x)=3sin(x)

prove

1 + cos (2 x) = 3 sin (x)

False

Solution steps

1 + cos (2 x) = 3 sin (x)

Show that the two sides are not equal

For

1 + cos (2 x) = 3 sin (x)

plug in

x = 0

1 + cos (2 \cdot 0) = 2

1 + cos (2 \cdot 0)

Simplify

= 1 + cos (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

= 1 + 1

Simplify

= 2

3 sin (0) = 0

3 sin (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

= 3 \cdot 0

Simplify

= 0

The two sides are not equal

\Rightarrow False

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\frac{1}{tan ( x )} = arctan (x)

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

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