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 Image Title*: Upload a Math Image The law of cosines is a trigonometric generalization of the Pythagorean Theorem. The law of cosines is a formula that helps in [[Solving Triangles|solving triangles]] when two or three side lengths of a triangle are known. Given a triangle with side lengths, $a,b,c$ and angle measures $A,B,C$, the law of cosines states ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ The formula combines the squares of two side lengths of a triangle and and a third term involving the [[cosine]] of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle. The law of cosines is useful in solving triangles whenever at least two side lengths are known. If the triangle that needs to be solved has less than just two known lengths, the [[Law of sines|law of sines]] is more useful. [[Image:Law of sines.jpg|right]] ===Alternate Forms=== When the law of cosines is being used to find a particular side length, that is when a certain angle measure is given, the law of cosines can be written in several different ways to help set up the equation to solve for the missing element of the triangle. :*$c^{2} = a^{2} + b^{2} - 2ab \cos C$ :*$b^{2} = a^{2} + c^{2} - 2ac \cos B$ :*$a^{2} = b^{2} + c^{2} - 2bc \cos A$ An alternate form of the law of cosines is particularly useful when solving a triangle when just the three side lengths are given. This form isolates the term with the cosine of the angle in it to make it easier to solve for the angle. :*$\cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}$ :*$\cos B = \frac{a^{2} + c^{2}- b^{2}}{2ac}$ :*$\cos A = \frac{b^{2} + c^{2}- a^{2}}{2bc}$ ==Proof== ===By the Pythagorean Theorem=== One way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle: ::$a^{2} + b^{2} = c^{2}$ [[Image:Pythagoras_law_of_cosines.jpg]] By Pythagorean theorem, we know ::$c^{2} = p^{2} + h^{2}$ But $p$ is just some portion of side length $a$ which is $r$ less than the length of $a$. Substituting the difference gives us, ::$c^{2} = (a-r)^{2} + h^{2}$ By Pythagorean theorem, we also know that ::$h^{2} = b^{2} - r^{2}$ Substituting the appropriate values gives us, ::$c^{2} = (a-r)^{2} + ( b^{2} - r^{2})$ Expanding the squared term gives us ::$c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}$ Simplify for ::$c^{2} = a^{2} -2ar + b^{2}$ And by the definition of cosine, we know that ::$r=b \cos C$ Substituting this value in give us ::$c^{2} = a^{2} -2ab \cos C + b^{2}$ or ::$c^{2} = a^{2}+ b^{2} -2ab \cos C$ ===Using the Distance Formula=== The law of cosines solves for a particular side length using the other side lengths and an angle. We can write this length using the distance formula as the distance from one vertex of the triangle to another. Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$. [[Image:Law_of_cosines_proof.jpg]] We use the formula for the distance between two points$(x_{1},y_{1}), (x_{2},y_{2})$ ::$\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ $c$ is the distance from $A$ to $B$. Since $A = (x_{1},y_{1}) = (b \cos C, b \sin C)$ and $B= (x_{2},y_{2})= (a,0)$, substituting the appropriate points into the distance formula gives us ::$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$ Squaring the inner terms, we have ::$c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$ Since $\cos^{2} C + \sin^{2} C = 1$, ::$c = \sqrt {a^{2}+b^{2}-2ab \cos C}$ Square both sides for ::$c^{2} = a^{2}+b^{2}-2ab \cos C$ ==Example Problem== Solve the triangle using the law of cosines. ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ [[Image:SAS triangle.jpg]] ===Solution=== {{hide|1= To find the side length $c$, :$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$ Simplify for ::$c^{2} =36 + 36 (2) - 72 \sqrt{2} \cos 45^\circ$ Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us ::$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$ Simplify for ::$c^{2} =36 + 72 - 72$ ::$c^{2} =36$ Taking the square root of both sides gives us ::$c =6$ ---- To solve for angle $B$, we can orient the triangle differently and use the law of cosines in the form [[Image:SAS_solution.jpg|right]] ::$b^{2} = a^{2} + c^{2} - 2ab \cos B$ Substituting in the appropriate side lengths gives us ::$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ ::$36 (2) = 36 + 36 - 72 \cos B$ ::$72 = 72 - 72 \cos B$ Subtracting $72$ from both sides gives us ::$0 = - 72 \cos B$ Dividing both sides by $-72$ gives us ::$0 = \cos B$ Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that ::$B = 90^\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$ ::$180^\circ - 90^\circ - 45^\circ = 45^\circ$ ::$A=45^\circ$ }} Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Topology Other All images were made by the page's author using Adobe Photoshop and Cinderella2. 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