What is the general solution for sin^2(a)-cos^2(a)=1 ?

The general solution for sin^2(a)-cos^2(a)=1 is a= pi/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(a)-cos^2(a)=1

sin^{2} (a) - cos^{2} (a) = 1

a = \frac{π}{2} + πn

Degrees

a = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (a) - cos^{2} (a) = 1

Rewrite using trig identities

sin^{2} (a) - cos^{2} (a)

Use the Double Angle identity:

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

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

= - cos (2 a)

- cos (2 a) = 1

Divide both sides by

- 1

- cos (2 a) = 1

Divide both sides by

- 1

\frac{- cos ( 2 a )}{- 1} = \frac{1}{- 1}

Simplify

cos (2 a) = - 1

cos (2 a) = - 1

General solutions for

cos (2 a) = - 1

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 a = π + 2 πn

2 a = π + 2 πn

Solve

2 a = π + 2 πn : a = \frac{π}{2} + πn

2 a = π + 2 πn

Divide both sides by

2

2 a = π + 2 πn

Divide both sides by

2

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

Simplify

a = \frac{π}{2} + πn

a = \frac{π}{2} + πn

a = \frac{π}{2} + πn

arcsin (x) = - 1

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

3 \cdot tan^{2} (x) = 1

2 \cdot 3 sin (v) + 2 cos (v) = 22

2 csc^{3} (x) + 21 csc^{2} (x) + 55 csc (x) + 42 = 0

What is the general solution for sin^2(a)-cos^2(a)=1 ?
The general solution for sin^2(a)-cos^2(a)=1 is a= pi/2+pin