What is sqrt(16-(4sin(θ))^2) ?

The solution to sqrt(16-(4sin(θ))^2) is 4sqrt(cos^2(θ))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify sqrt(16-(4sin(θ))^2)

simplify

16 - (4 sin (θ))^{2}

4 cos^{2} (θ)

Solution steps

16 - (4 sin (θ))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 16 - 4^{2} sin^{2} (θ)

4^{2} = 16

= 16 - 16 sin^{2} (θ)

Factor

16 - 16 sin^{2} (θ) : 16 (1 - sin^{2} (θ))

= 16 (1 - sin^{2} (θ))

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 16 - sin^{2} (θ) + 1

16 = 4

= 4 - sin^{2} (θ) + 1

Rewrite using trig identities

= 4 cos^{2} (θ)

csc (arccos (u))

cot (x - π)

1 - (cos (x))^{2}

cos (- x) - sin (- x)

tan (π + θ)

What is sqrt(16-(4sin(θ))^2) ?
The solution to sqrt(16-(4sin(θ))^2) is 4sqrt(cos^2(θ))