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Popular >

sin^2(1/4)+cos^2(a)=1

Solution

sin^{2} (\frac{1}{4}) + cos^{2} (a) = 1

Solution

a = 0.25 + 2 πn, a = 2 π - 0.25 + 2 πn, a = 2.89159 \dots + 2 πn, a = - 2.89159 \dots + 2 πn

+1

Degrees

a = 14.32394 \dots^{\circ} + 36 0^{\circ} n, a = 345.67605 \dots^{\circ} + 36 0^{\circ} n, a = 165.67605 \dots^{\circ} + 36 0^{\circ} n, a = - 165.67605 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (\frac{1}{4}) + cos^{2} (a) = 1

Solve by substitution

sin^{2} (\frac{1}{4}) + cos^{2} (a) = 1

Let:

cos (a) = u

sin^{2} (\frac{1}{4}) + u^{2} = 1

sin^{2} (\frac{1}{4}) + u^{2} = 1 : u = 1 - sin^{2} (\frac{1}{4}), u = - 1 - sin^{2} (\frac{1}{4})

sin^{2} (\frac{1}{4}) + u^{2} = 1

Move

sin^{2} (\frac{1}{4})

to the right side

sin^{2} (\frac{1}{4}) + u^{2} = 1

Subtract

sin^{2} (\frac{1}{4})

from both sides

sin^{2} (\frac{1}{4}) + u^{2} - sin^{2} (\frac{1}{4}) = 1 - sin^{2} (\frac{1}{4})

Simplify

u^{2} = 1 - sin^{2} (\frac{1}{4})

u^{2} = 1 - sin^{2} (\frac{1}{4})

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1 - sin^{2} (\frac{1}{4}), u = - 1 - sin^{2} (\frac{1}{4})

Substitute back

u = cos (a)

cos (a) = 1 - sin^{2} (\frac{1}{4}), cos (a) = - 1 - sin^{2} (\frac{1}{4})

cos (a) = 1 - sin^{2} (\frac{1}{4}), cos (a) = - 1 - sin^{2} (\frac{1}{4})

cos (a) = 1 - sin^{2} (\frac{1}{4}) : a = arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn, a = 2 π - arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn

cos (a) = 1 - sin^{2} (\frac{1}{4})

Apply trig inverse properties

cos (a) = 1 - sin^{2} (\frac{1}{4})

General solutions for

cos (a) = 1 - sin^{2} (\frac{1}{4})

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

a = arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn, a = 2 π - arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn

a = arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn, a = 2 π - arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn

cos (a) = - 1 - sin^{2} (\frac{1}{4}) : a = arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn, a = - arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn

cos (a) = - 1 - sin^{2} (\frac{1}{4})

Apply trig inverse properties

cos (a) = - 1 - sin^{2} (\frac{1}{4})

General solutions for

cos (a) = - 1 - sin^{2} (\frac{1}{4})

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

a = arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn, a = - arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn

a = arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn, a = - arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn

Combine all the solutions

a = arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn, a = 2 π - arccos (1 - sin^{2} (\frac{1}{4})) + 2 πn, a = arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn, a = - arccos (- 1 - sin^{2} (\frac{1}{4})) + 2 πn

Show solutions in decimal form

a = 0.25 + 2 πn, a = 2 π - 0.25 + 2 πn, a = 2.89159 \dots + 2 πn, a = - 2.89159 \dots + 2 πn

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