prove sin(x)= 5/13

Solution

prove

sin (x) = \frac{5}{13}

False

Solution steps

sin (x) = \frac{5}{13}

Show that the two sides are not equal

For

sin (x) = \frac{5}{13}

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

The two sides are not equal

\Rightarrow False

sec (x) - tan (x) \cdot sin (x) = \frac{1}{sec ( x )}

sin^{4} (x) + cos^{4} (x) = 1 - \frac{1}{2} sin^{2} (2 x)

\frac{( 1 + cos ( 2 y ) )}{sin ^{2} ( y )} = 2 cot^{2} (y)

- tan (t) - \frac{cos ( t )}{sin ( t ) - 1} = sec (t)

4 cos^{2} (B) - 4 sin^{2} (B) = 4 - 8 sin^{2} (B)