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Popular >

solvefor x,sin(x+y)+sin(x-y)=1

Solution

solve for

x, sin (x + y) + sin (x - y) = 1

Solution

x = arcsin (\frac{1}{2 cos ( y )}) + 2 πn, x = π + arcsin (- \frac{1}{2 cos ( y )}) + 2 πn

Solution steps

sin (x + y) + sin (x - y) = 1

Rewrite using trig identities

sin (x + y) + sin (x - y)

Use the Sum to Product identity:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

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

Simplify

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

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

\frac{x + y + x - y}{2} = x

\frac{x + y + x - y}{2}

x + y + x - y = 2 x

x + y + x - y

Group like terms

= y - y + x + x

Add similar elements:

y - y = 0

= x + x

Add similar elements:

x + x = 2 x

= 2 x

= \frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

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

\frac{x + y - ( x - y )}{2}

Expand

x + y - (x - y) : 2 y

x + y - (x - y)

- (x - y) : - x + y

- (x - y)

Distribute parentheses

= - x - (- y)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x + y

= x + y - x + y

Simplify

x + y - x + y : 2 y

x + y - x + y

Group like terms

= y + y + x - x

Add similar elements:

y + y = 2 y

= 2 y + x - x

Add similar elements:

x - x = 0

= 2 y

= 2 y

= \frac{2 y}{2}

Divide the numbers:

\frac{2}{2} = 1

= y

= 2 cos (y) sin (x)

= 2 cos (y) sin (x)

2 cos (y) sin (x) = 1

Divide both sides by

2 cos (y); y \neq = \frac{π}{2} + 2 πn, y \neq = \frac{3 π}{2} + 2 πn

2 cos (y) sin (x) = 1

Divide both sides by

2 cos (y); y \neq = \frac{π}{2} + 2 πn, y \neq = \frac{3 π}{2} + 2 πn

\frac{2 cos ( y ) sin ( x )}{2 cos ( y )} = \frac{1}{2 cos ( y )}; y \neq = \frac{π}{2} + 2 πn, y \neq = \frac{3 π}{2} + 2 πn

Simplify

sin (x) = \frac{1}{2 cos ( y )}; y \neq = \frac{π}{2} + 2 πn, y \neq = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2 cos ( y )}; y \neq = \frac{π}{2} + 2 πn, y \neq = \frac{3 π}{2} + 2 πn

Apply trig inverse properties

sin (x) = \frac{1}{2 cos ( y )}

General solutions for

sin (x) = \frac{1}{2 cos ( y )}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{1}{2 cos ( y )}) + 2 πn, x = π + arcsin (- \frac{1}{2 cos ( y )}) + 2 πn

x = arcsin (\frac{1}{2 cos ( y )}) + 2 πn, x = π + arcsin (- \frac{1}{2 cos ( y )}) + 2 πn

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