Is prove-50cos(4t)+50=100sin^2(2t) ?

The answer to whether prove-50cos(4t)+50=100sin^2(2t) is True

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prove-50cos(4t)+50=100sin^2(2t)

prove

- 50 cos (4 t) + 50 = 100 sin^{2} (2 t)

True

Solution steps

- 50 cos (4 t) + 50 = 100 sin^{2} (2 t)

Manipulating left side

- 50 cos (4 t) + 50

Rewrite using trig identities

- 50 cos (4 t) + 50

= - 50 cos (2 \cdot 2 t) + 50

Use the Double Angle identity:

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

= - 50 (1 - 2 sin^{2} (2 t)) + 50

Simplify

- 50 (1 - 2 sin^{2} (2 t)) + 50 : 100 sin^{2} (2 t)

- 50 (1 - 2 sin^{2} (2 t)) + 50

Expand

- 50 (1 - 2 sin^{2} (2 t)) : - 50 + 100 sin^{2} (2 t)

- 50 (1 - 2 sin^{2} (2 t))

Apply the distributive law:

a (b - c) = ab - a c

a = - 50, b = 1, c = 2 sin^{2} (2 t)

= - 50 \cdot 1 - (- 50) \cdot 2 sin^{2} (2 t)

Apply minus-plus rules

- (- a) = a

= - 50 \cdot 1 + 50 \cdot 2 sin^{2} (2 t)

Simplify

- 50 \cdot 1 + 50 \cdot 2 sin^{2} (2 t) : - 50 + 100 sin^{2} (2 t)

- 50 \cdot 1 + 50 \cdot 2 sin^{2} (2 t)

Multiply the numbers:

50 \cdot 1 = 50

= - 50 + 50 \cdot 2 sin^{2} (2 t)

Multiply the numbers:

50 \cdot 2 = 100

= - 50 + 100 sin^{2} (2 t)

= - 50 + 100 sin^{2} (2 t)

= - 50 + 100 sin^{2} (2 t) + 50

Simplify

- 50 + 100 sin^{2} (2 t) + 50 : 100 sin^{2} (2 t)

- 50 + 100 sin^{2} (2 t) + 50

Group like terms

= 100 sin^{2} (2 t) - 50 + 50

- 50 + 50 = 0

= 100 sin^{2} (2 t)

= 100 sin^{2} (2 t)

= 100 sin^{2} (2 t)

= 100 sin^{2} (2 t)

We showed that the two sides could take the same form

\Rightarrow True

\frac{7 sec ( θ ) - 7}{1 - cos ( θ )} = 7 sec (θ)

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

cos (20 π + x) + cos (21 π - x) = 0

\frac{( tan ^{2} ( x ) + 1 )}{sec ( x )} = sec (x)

csc^{2} (x) cos (2 x) = 1 - tan^{2} (x)

Is prove-50cos(4t)+50=100sin^2(2t) ?
The answer to whether prove-50cos(4t)+50=100sin^2(2t) is True