Is sin^2(90-b)+cos^2(b-450)=1 ?

The answer to whether sin^2(90-b)+cos^2(b-450)=1 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin^2(90-b)+cos^2(b-450)=1

Solution

prove

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

Solution

True

Solution steps

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

Manipulating left side

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ})

Rewrite using trig identities

cos (b - 45 0^{\circ})

Use the Angle Difference identity:

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

= cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

Simplify

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ}) : sin (b)

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

cos (b) cos (45 0^{\circ}) = 0

cos (b) cos (45 0^{\circ})

cos (45 0^{\circ}) = 0

cos (45 0^{\circ})

cos (45 0^{\circ}) = cos (9 0^{\circ})

cos (45 0^{\circ})

Rewrite

45 0^{\circ}

36 0^{\circ} + 9 0^{\circ}

= cos (36 0^{\circ} + 9 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ}) = cos (x)

cos (36 0^{\circ} + 9 0^{\circ}) = cos (9 0^{\circ})

= cos (9 0^{\circ})

= cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (9 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0

= 0 \cdot cos (b)

Apply rule

0 \cdot a = 0

= 0

sin (b) sin (45 0^{\circ}) = sin (b)

sin (b) sin (45 0^{\circ})

sin (45 0^{\circ}) = 1

sin (45 0^{\circ})

sin (45 0^{\circ}) = sin (9 0^{\circ})

sin (45 0^{\circ})

Rewrite

45 0^{\circ}

36 0^{\circ} + 9 0^{\circ}

= sin (36 0^{\circ} + 9 0^{\circ})

Apply the periodicity of

sin

sin (x + 36 0^{\circ}) = sin (x)

sin (36 0^{\circ} + 9 0^{\circ}) = sin (9 0^{\circ})

= sin (9 0^{\circ})

= sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1

= 1 \cdot sin (b)

Multiply:

sin (b) \cdot 1 = sin (b)

= sin (b)

= 0 + sin (b)

0 + sin (b) = sin (b)

= sin (b)

= sin (b)

= sin^{2} (9 0^{\circ} - b) + (sin (b))^{2}

Rewrite using trig identities

sin (9 0^{\circ} - b)

Use the Angle Difference identity:

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

= sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b)

Simplify

sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b) : cos (b)

sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b)

sin (9 0^{\circ}) cos (b) = cos (b)

sin (9 0^{\circ}) cos (b)

Simplify

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 cos (b)

Multiply:

1 \cdot cos (b) = cos (b)

= cos (b)

cos (9 0^{\circ}) sin (b) = 0

cos (9 0^{\circ}) sin (b)

Simplify

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 sin (b)

Apply rule

0 \cdot a = 0

= 0

= cos (b) - 0

cos (b) - 0 = cos (b)

= cos (b)

= cos (b)

= (cos (b))^{2} + (sin (b))^{2}

Remove parentheses:

(a) = a

= cos^{2} (b) + sin^{2} (b)

Rewrite using trig identities

cos^{2} (b) + sin^{2} (b)

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

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

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

cos^2(x)=(1-tan^2(x))/(sec^2(x))

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

sin^2(x)-3cos(x)-4=0

sin^{2} (x) - 3 cos (x) - 4 = 0

tan^2(x)+sec(x)=5

tan^{2} (x) + sec (x) = 5

4sin^5(x)=3

4 sin^{5} (x) = 3

Frequently Asked Questions (FAQ)

Is sin^2(90-b)+cos^2(b-450)=1 ?
The answer to whether sin^2(90-b)+cos^2(b-450)=1 is True