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Popular Trigonometry >

a^2=b^2+c^2-2bc*cos(x)

Solution

a^{2} = b^{2} + c^{2} - 2 b c \cdot cos (x)

Solution

x = arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn, x = - arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn

Solution steps

a^{2} = b^{2} + c^{2} - 2 b c cos (x)

Switch sides

b^{2} + c^{2} - 2 b c cos (x) = a^{2}

Move

b^{2}

to the right side

b^{2} + c^{2} - 2 b c cos (x) = a^{2}

Subtract

b^{2}

from both sides

b^{2} + c^{2} - 2 b c cos (x) - b^{2} = a^{2} - b^{2}

Simplify

c^{2} - 2 b c cos (x) = a^{2} - b^{2}

c^{2} - 2 b c cos (x) = a^{2} - b^{2}

Move

c^{2}

to the right side

c^{2} - 2 b c cos (x) = a^{2} - b^{2}

Subtract

c^{2}

from both sides

c^{2} - 2 b c cos (x) - c^{2} = a^{2} - b^{2} - c^{2}

Simplify

- 2 b c cos (x) = a^{2} - b^{2} - c^{2}

- 2 b c cos (x) = a^{2} - b^{2} - c^{2}

Divide both sides by

- 2 b c; - 2 b c \neq = 0

- 2 b c cos (x) = a^{2} - b^{2} - c^{2}

Divide both sides by

- 2 b c; - 2 b c \neq = 0

\frac{- 2 b c cos ( x )}{- 2 b c} = \frac{a ^{2}}{- 2 b c} - \frac{b ^{2}}{- 2 b c} - \frac{c ^{2}}{- 2 b c}; - 2 b c \neq = 0

Simplify

\frac{- 2 b c cos ( x )}{- 2 b c} = \frac{a ^{2}}{- 2 b c} - \frac{b ^{2}}{- 2 b c} - \frac{c ^{2}}{- 2 b c}

Simplify

\frac{- 2 b c cos ( x )}{- 2 b c} : cos (x)

\frac{- 2 b c cos ( x )}{- 2 b c}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 b c cos ( x )}{2 b c}

Divide the numbers:

\frac{2}{2} = 1

= \frac{b c cos ( x )}{b c}

Cancel the common factor:

b

= \frac{c cos ( x )}{c}

Cancel the common factor:

c

= cos (x)

Simplify

\frac{a ^{2}}{- 2 b c} - \frac{b ^{2}}{- 2 b c} - \frac{c ^{2}}{- 2 b c} : - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}

\frac{a ^{2}}{- 2 b c} - \frac{b ^{2}}{- 2 b c} - \frac{c ^{2}}{- 2 b c}

Apply rule

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

= \frac{a ^{2} - b ^{2} - c ^{2}}{- 2 b c}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}

cos (x) = - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}; - 2 b c \neq = 0

cos (x) = - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}; - 2 b c \neq = 0

cos (x) = - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}; - 2 b c \neq = 0

Apply trig inverse properties

cos (x) = - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}

General solutions for

cos (x) = - \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}

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

x = arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn, x = - arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn

x = arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn, x = - arccos (- \frac{a ^{2} - b ^{2} - c ^{2}}{2 b c}) + 2 πn

Graph

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