Is sqrt(81-9(3sin(x))^2)=-1 ?

The answer to whether sqrt(81-9(3sin(x))^2)=-1 is False

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prove sqrt(81-9(3sin(x))^2)=-1

prove

81 - 9 (3 sin (x))^{2} = - 1

False

Solution steps

81 - 9 (3 sin (x))^{2} = - 1

Show that the two sides are not equal

For

81 - 9 (3 sin (x))^{2} = - 1

plug in

x = 0

81 - 9 (3 sin (0))^{2} = 9

81 - 9 (3 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

= 81 - 9 (3 \cdot 0)^{2}

Simplify

81 - 9 (3 \cdot 0)^{2} : 9

81 - 9 (3 \cdot 0)^{2}

9 (3 \cdot 0)^{2} = 0

9 (3 \cdot 0)^{2}

(3 \cdot 0)^{2} = 0

(3 \cdot 0)^{2}

Apply rule

0 \cdot a = 0

= 0^{2}

Apply rule

0^{a} = 0

= 0

= 9 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 81 - 0

Subtract the numbers:

81 - 0 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

n a^{n} = a

9^{2} = 9

= 9

= 9

The two sides are not equal

\Rightarrow False

p ro v e csc (x) + 1 = \frac{2}{sin ( x )}

p ro v e \frac{1}{cos ( x )} = arccos (x)

p ro v e tan^{2} (x) sec (x) = cot^{2} (x) csc (x)

p ro v e (sec^{2} (x) - 1) csc (x) = tan^{2} (x) csc (x)

p ro v e 1 - sec (x) = \frac{( cos ( x ) - 1 )}{cos ( x )}

Is sqrt(81-9(3sin(x))^2)=-1 ?
The answer to whether sqrt(81-9(3sin(x))^2)=-1 is False