Is cos^2(b)+cos^2(pi/2-b)=1 ?

The answer to whether cos^2(b)+cos^2(pi/2-b)=1 is True

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prove cos^2(b)+cos^2(pi/2-b)=1

prove

cos^{2} (b) + cos^{2} (\frac{π}{2} - b) = 1

True

Solution steps

cos^{2} (b) + cos^{2} (\frac{π}{2} - b) = 1

Manipulating left side

cos^{2} (b) + cos^{2} (\frac{π}{2} - b)

Rewrite using trig identities

cos (\frac{π}{2} - b)

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= cos (\frac{π}{2}) cos (b) + sin (\frac{π}{2}) sin (b)

Simplify

cos (\frac{π}{2}) cos (b) + sin (\frac{π}{2}) sin (b) : sin (b)

cos (\frac{π}{2}) cos (b) + sin (\frac{π}{2}) sin (b)

cos (\frac{π}{2}) cos (b) = 0

cos (\frac{π}{2}) cos (b)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (b)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (b) = sin (b)

sin (\frac{π}{2}) sin (b)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot sin (b)

Multiply:

1 \cdot sin (b) = sin (b)

= sin (b)

= 0 + sin (b)

0 + sin (b) = sin (b)

= sin (b)

= sin (b)

= cos^{2} (b) + (sin (b))^{2}

Remove parentheses:

(a) = a

= cos^{2} (b) + sin^{2} (b)

Rewrite using trig identities

cos^{2} (b) + sin^{2} (b)

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sec (x) - (sin (x)) (tan (x)) = cos (x)

\frac{1}{tan ( θ )} + tan (θ) = sec (θ) csc (θ)

sec (x) - \frac{1}{sec ( x )} = sin (x) tan (x)

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )} = sec (b)

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

Is cos^2(b)+cos^2(pi/2-b)=1 ?
The answer to whether cos^2(b)+cos^2(pi/2-b)=1 is True