Is (5sin(θ)-5sin(θ))^2=25-25sin(2θ) ?

The answer to whether (5sin(θ)-5sin(θ))^2=25-25sin(2θ) is False

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prove (5sin(θ)-5sin(θ))^2=25-25sin(2θ)

prove

(5 sin (θ) - 5 sin (θ))^{2} = 25 - 25 sin (2 θ)

False

Solution steps

(5 sin (θ) - 5 sin (θ))^{2} = 25 - 25 sin (2 θ)

Show that the two sides are not equal

For

(5 sin (θ) - 5 sin (θ))^{2} = 25 - 25 sin (2 θ)

plug in

θ = 0

(5 sin (0) - 5 sin (0))^{2} = 0

(5 sin (0) - 5 sin (0))^{2}

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

= (5 \cdot 0 - 5 \cdot 0)^{2}

Simplify

= 0

25 - 25 sin (2 \cdot 0) = 25

25 - 25 sin (2 \cdot 0)

Simplify

= 25 - 25 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

= 25 - 25 \cdot 0

Simplify

= 25

The two sides are not equal

\Rightarrow False

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

\frac{1}{( sec ( y ) - tan ( y ) )} = sec (y) + tan (y)

tan (θ) (cos (θ) + cot (θ)) = sin (θ) + 1

2 cos (2 x) = 4 cos^{2} (x) - 2

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

Is (5sin(θ)-5sin(θ))^2=25-25sin(2θ) ?
The answer to whether (5sin(θ)-5sin(θ))^2=25-25sin(2θ) is False