What is the general solution for cos^2(φ)=sin^2(φ) ?

The general solution for cos^2(φ)=sin^2(φ) is φ= pi/4+pin,φ=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(φ)=sin^2(φ)

Solution

cos^{2} (φ) = sin^{2} (φ)

Solution

φ = \frac{π}{4} + πn, φ = \frac{3 π}{4} + πn

Degrees

φ = 4 5^{\circ} + 18 0^{\circ} n, φ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (φ) = sin^{2} (φ)

Subtract

sin^{2} (φ)

from both sides

cos^{2} (φ) - sin^{2} (φ) = 0

Rewrite using trig identities

cos^{2} (φ) - sin^{2} (φ)

Use the Double Angle identity:

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

= cos (2 φ)

cos (2 φ) = 0

General solutions for

cos (2 φ) = 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 φ = \frac{π}{2} + 2 πn, 2 φ = \frac{3 π}{2} + 2 πn

2 φ = \frac{π}{2} + 2 πn, 2 φ = \frac{3 π}{2} + 2 πn

Solve

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

2 φ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 φ = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 φ}{2} : φ

\frac{2 φ}{2}

Divide the numbers:

\frac{2}{2} = 1

= φ

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

φ = \frac{π}{4} + πn

φ = \frac{π}{4} + πn

φ = \frac{π}{4} + πn

Solve

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

2 φ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 φ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 φ}{2} : φ

\frac{2 φ}{2}

Divide the numbers:

\frac{2}{2} = 1

= φ

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

φ = \frac{3 π}{4} + πn

φ = \frac{3 π}{4} + πn

φ = \frac{3 π}{4} + πn

φ = \frac{π}{4} + πn, φ = \frac{3 π}{4} + πn

Popular Examples

sin^2(x)-2cos^2(x)=0

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

sin(x)= 1/(sqrt(10))

sin (x) = \frac{1}{10}

(sin(θ)-1)(sin(θ)+1)=0

(sin (θ) - 1) (sin (θ) + 1) = 0

5sec(θ)-6=0

5 sec (θ) - 6 = 0

sec(x)=-1,-2pi<= x<= 2pi

sec (x) = - 1, - 2 π \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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