Is sin(θ)cos(θ)=1 ?

The answer to whether sin(θ)cos(θ)=1 is False

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prove sin(θ)cos(θ)=1

prove

sin (θ) cos (θ) = 1

False

Solution steps

sin (θ) cos (θ) = 1

Show that the two sides are not equal

For

sin (θ) cos (θ) = 1

plug in

θ = 0

sin (0) cos (0) = 0

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

= 0 \cdot 1

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

csc (θ) - csc (θ) cos^{2} (θ) = sin (θ)

\frac{sin ( 6 x ) + sin ( 2 x )}{cos ( 6 x ) + cos ( 2 x )} = tan (4 x)

sec^{2} (a) = 1 + tan^{2} (a)

\frac{tan ( θ )}{csc ( θ )} = sec (θ) - cos (θ)