Is prove-2=-2cos(0)+15sin(3)(0) ?

The answer to whether prove-2=-2cos(0)+15sin(3)(0) is True

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prove-2=-2cos(0)+15sin(3)(0)

prove

- 2 = - 2 cos (0) + 15 sin (3) (0)

True

Solution steps

- 2 = - 2 cos (0) + 15 sin (3) \cdot 0

Manipulating right side

- 2 cos (0) + 15 sin (3) 0

Simplify

- 2 cos (0) + 0 \cdot 15 sin (3) : - 2

- 2 cos (0) + 0 \cdot 15 sin (3)

2 cos (0) = 2

2 cos (0)

Simplify

cos (0) : 1

cos (0)

Use the following trivial identity:

cos (0) = 1

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

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

0 \cdot 15 sin (3) = 0

0 \cdot 15 sin (3)

Apply rule

0 \cdot a = 0

= 0

= - 2 + 0

Simplify

= - 2

= - 2

We showed that the two sides could take the same form

\Rightarrow True

tan (x) - sin (x) = tan (x) sin (x)

tan (x) cot^{2} (x) = \frac{cos ( x )}{sin ( x )}

sin (x) + \frac{cos ( x )}{tan ( x )} = csc (x)

\frac{1}{sin ^{2} ( θ )} - \frac{1}{tan ^{2} ( θ )} = 1

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

Is prove-2=-2cos(0)+15sin(3)(0) ?
The answer to whether prove-2=-2cos(0)+15sin(3)(0) is True