Is sin(x+y)=cos(pi/2-(x+y)) ?

The answer to whether sin(x+y)=cos(pi/2-(x+y)) is True

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prove sin(x+y)=cos(pi/2-(x+y))

prove

sin (x + y) = cos (\frac{π}{2} - (x + y))

True

Solution steps

sin (x + y) = cos (\frac{π}{2} - (x + y))

Manipulating right side

cos (\frac{π}{2} - (x + y))

Rewrite using trig identities

cos (\frac{π}{2} - (x + y))

Use the Angle Difference identity:

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

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

Simplify

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

cos (\frac{π}{2}) cos (x + y) + sin (\frac{π}{2}) sin (x + y)

cos (\frac{π}{2}) cos (x + y) = 0

cos (\frac{π}{2}) cos (x + y)

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 (x + y)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x + y) = sin (x + y)

sin (\frac{π}{2}) sin (x + y)

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 (x + y)

Multiply:

1 \cdot sin (x + y) = sin (x + y)

= sin (x + y)

= 0 + sin (x + y)

0 + sin (x + y) = sin (x + y)

= sin (x + y)

= sin (x + y)

= sin (x + y)

We showed that the two sides could take the same form

\Rightarrow True

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

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

csc (s) - sin (s) = cos (s) cot (s)

\frac{( 4 sin ^{2} ( x ) )}{tan ^{2} ( x )} = 4 cos^{2} (x)

\frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )} = tan (3 x)

Is sin(x+y)=cos(pi/2-(x+y)) ?
The answer to whether sin(x+y)=cos(pi/2-(x+y)) is True