What is the range of sin^2(θ) ?

The range of sin^2(θ) is 0<= f(θ)<= 1

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range of sin^2(θ)

range

sin^{2} (θ)

0 \leq f (θ) \leq 1

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

sin^{2} (θ) : - \infty < θ < \infty

Extreme Points of

sin^{2} (θ) :

Minimum

(πn, 0),

Maximum

(\frac{π}{2} + πn, 1)

Find the range for the interval

- \infty < θ < \infty : 0 \leq f (θ) \leq 1

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (θ) \leq 1

f (x) = 2 x^{2} - 7

f (x) = 36 - x^{2} + x + 2

- 2 x + y = - 4

f (x) = x^{2} - 4

\frac{6 x + 3}{x + 4}