What is the value of sin^2(pi)+cos^2(pi) ?

The value of sin^2(pi)+cos^2(pi) is 1

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sin^2(pi)+cos^2(pi)

sin^{2} (π) + cos^{2} (π)

1

Solution steps

sin^{2} (π) + cos^{2} (π)

Use the following trivial identity:

sin (π) = 0

sin (π)

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 (π) = (- 1)

cos (π)

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^{2} + (- 1)^{2}

Simplify

0^{2} + (- 1)^{2} : 1

0^{2} + (- 1)^{2}

Apply rule

0^{a} = 0

0^{2} = 0

= 0 + (- 1)^{2}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 0 + 1

Add the numbers:

0 + 1 = 1

= 1

= 1

\frac{1}{cos ^{2} ( \frac{π}{6} )}

\frac{16.58}{2} \cdot sin (4^{\circ})

sin (0 + π)

0.15 cos (2 π \cdot 0.25)

24 sin (4 5^{\circ})

What is the value of sin^2(pi)+cos^2(pi) ?
The value of sin^2(pi)+cos^2(pi) is 1