What is the value of cos(2arcsin(3/5)) ?

The value of cos(2arcsin(3/5)) is 7/25

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cos(2arcsin(3/5))

cos (2 arcsin (\frac{3}{5}))

\frac{7}{25}

Decimal Notation

0.28

Solution steps

cos (2 arcsin (\frac{3}{5}))

Rewrite using trig identities:

1 - 2 sin^{2} (arcsin (\frac{3}{5}))

cos (2 arcsin (\frac{3}{5}))

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (arcsin (\frac{3}{5}))

= 1 - 2 sin^{2} (arcsin (\frac{3}{5}))

Rewrite using trig identities:

sin (arcsin (\frac{3}{5})) = \frac{3}{5}

Use the following identity:

sin (arcsin (x)) = x

= \frac{3}{5}

= 1 - 2 (\frac{3}{5})^{2}

Simplify

1 - 2 (\frac{3}{5})^{2} : \frac{7}{25}

1 - 2 (\frac{3}{5})^{2}

2 (\frac{3}{5})^{2} = \frac{18}{25}

2 (\frac{3}{5})^{2}

(\frac{3}{5})^{2} = \frac{3 ^{2}}{5 ^{2}}

(\frac{3}{5})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{3 ^{2}}{5 ^{2}}

= 2 \cdot \frac{3 ^{2}}{5 ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ^{2} \cdot 2}{5 ^{2}}

3^{2} \cdot 2 = 18

3^{2} \cdot 2

3^{2} = 9

= 9 \cdot 2

Multiply the numbers:

9 \cdot 2 = 18

= 18

= \frac{18}{5 ^{2}}

5^{2} = 25

= \frac{18}{25}

= 1 - \frac{18}{25}

Convert element to fraction:

1 = \frac{1 \cdot 25}{25}

= \frac{1 \cdot 25}{25} - \frac{18}{25}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 25 - 18}{25}

1 \cdot 25 - 18 = 7

1 \cdot 25 - 18

Multiply the numbers:

1 \cdot 25 = 25

= 25 - 18

Subtract the numbers:

25 - 18 = 7

= 7

= \frac{7}{25}

= \frac{7}{25}

tan (- 33 0^{\circ})

arctan (e^{0})

tan (4 π)

arccos (- \frac{4}{5})

tan (- 22 5^{\circ})