Is 9sin(2x)sin(x)=9cos(x) ?

The answer to whether 9sin(2x)sin(x)=9cos(x) is False

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prove 9sin(2x)sin(x)=9cos(x)

prove

9 sin (2 x) sin (x) = 9 cos (x)

False

Solution steps

9 sin (2 x) sin (x) = 9 cos (x)

Show that the two sides are not equal

For

9 sin (2 x) sin (x) = 9 cos (x)

plug in

x = 0

9 sin (2 \cdot 0) sin (0) = 0

9 sin (2 \cdot 0) sin (0)

Simplify

= 9 sin (0) 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

= 9 \cdot 0 \cdot 0

Simplify

= 0

9 cos (0) = 9

9 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

= 9 \cdot 1

Simplify

= 9

The two sides are not equal

\Rightarrow False

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

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Is 9sin(2x)sin(x)=9cos(x) ?
The answer to whether 9sin(2x)sin(x)=9cos(x) is False