The answer to whether sin(x)=5cos(x) is False

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prove sin(x)=5cos(x)

Solution

prove

sin (x) = 5 cos (x)

False

Solution steps

sin (x) = 5 cos (x)

Show that the two sides are not equal

For

sin (x) = 5 cos (x)

plug in

x = 0

sin (0) = 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

= 0

5 cos (0) = 5

5 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

= 5 \cdot 1

Simplify

= 5

The two sides are not equal

\Rightarrow False

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

p ro v e \frac{1}{sin ( x ) - sin ( x ) cos ( x )} = csc (x)

p ro v e sin (a) csc (a) = 1

p ro v e cos^{2} (θ) = \frac{( 1 + cos ( 2 θ ) )}{2}

p ro v e cos (33 0^{\circ}) = 1 - 2 sin^{2} (16 5^{\circ})