What is the general solution for solvefor x,cos^2(x+y)=sin^2(x+y) ?

The general solution for solvefor x,cos^2(x+y)=sin^2(x+y) is x= pi/4+pin-y,x=(3pi)/4+pin-y

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solvefor x,cos^2(x+y)=sin^2(x+y)

Solution

solve for

x, cos^{2} (x + y) = sin^{2} (x + y)

Solution

x = \frac{π}{4} + πn - y, x = \frac{3 π}{4} + πn - y

Solution steps

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

Subtract

sin^{2} (x + y)

from both sides

cos^{2} (x + y) - sin^{2} (x + y) = 0

Rewrite using trig identities

cos^{2} (x + y) - sin^{2} (x + y)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= cos (2 (y + x))

cos (2 (y + x)) = 0

General solutions for

cos (2 (y + x)) = 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}

2 (y + x) = \frac{π}{2} + 2 πn, 2 (y + x) = \frac{3 π}{2} + 2 πn

2 (y + x) = \frac{π}{2} + 2 πn, 2 (y + x) = \frac{3 π}{2} + 2 πn

Solve

2 (y + x) = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn - y

2 (y + x) = \frac{π}{2} + 2 πn

Divide both sides by

2

2 (y + x) = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

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

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

Divide the numbers:

\frac{2}{2} = 1

= y + x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

y + x = \frac{π}{4} + πn

y + x = \frac{π}{4} + πn

y + x = \frac{π}{4} + πn

Move

y

to the right side

y + x = \frac{π}{4} + πn

Subtract

y

from both sides

y + x - y = \frac{π}{4} + πn - y

Simplify

x = \frac{π}{4} + πn - y

x = \frac{π}{4} + πn - y

Solve

2 (y + x) = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn - y

2 (y + x) = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 (y + x) = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 ( y + x )}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 ( y + x )}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

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

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

Divide the numbers:

\frac{2}{2} = 1

= y + x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

y + x = \frac{3 π}{4} + πn

y + x = \frac{3 π}{4} + πn

y + x = \frac{3 π}{4} + πn

Move

y

to the right side

y + x = \frac{3 π}{4} + πn

Subtract

y

from both sides

y + x - y = \frac{3 π}{4} + πn - y

Simplify

x = \frac{3 π}{4} + πn - y

x = \frac{3 π}{4} + πn - y

x = \frac{π}{4} + πn - y, x = \frac{3 π}{4} + πn - y

Popular Examples

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cot^{2} (x) + 3 cot (x) + 2 = cot (x) + 1

a^2=b^2+c^2-2bc*cos(x)

a^{2} = b^{2} + c^{2} - 2 b c \cdot cos (x)

3sin^2(X)-cos(X)=0

3 sin^{2} (X) - cos (X) = 0

4sin(θ)=csc(θ)

4 sin (θ) = csc (θ)

-1/3 =sin(x)

- \frac{1}{3} = sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,cos^2(x+y)=sin^2(x+y) ?
The general solution for solvefor x,cos^2(x+y)=sin^2(x+y) is x= pi/4+pin-y,x=(3pi)/4+pin-y