Date modified: Mon 2023-09-18 . 09:01 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Trigonometry, law of cosine and the law of sine Let us review some basic trigonometric definitions by the following diagram, recall one way to define them is that they are **ratios of lengths** in a right triangle: ![[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.09.19.excalidraw.svg]] %%[[1 teaching/summer program 2023/week 1/---files/Trigonometry_cosine_sine 2023-05-10 08.09.19.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 1/---files/Trigonometry_cosine_sine 2023-05-10 08.09.19.excalidraw.dark.svg|dark exported image]]%% This gives the trigonometric Pythagorean theorem: $$\cos^{2}\theta+\sin^{2}\theta=1$$ ## Using yarn to measure the height of a mountain. Take a piece of yarn and tape it to the side of a sheet of paper. Can you measure the height of the ceiling in this classroom, or a height of a mountain? ![[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.20.32.excalidraw.svg]] %%[[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.20.32.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 1/---files/Trigonometry_cosine_sine 2023-05-10 08.20.32.excalidraw.dark.svg|dark exported image]]%% --- Given a generic triangle with sides a,b,c, what are their relations if they are not necessarily a right triangle? In this case we have the **law of cosine**, which is Pythagorean theorem plus a correction term. >**Theorem. Law of cosine.** >Given a triangle with side lengths $a,b,c$, and let $\gamma$ be the angle between sides $a$ and $b$. Then $a^{2}+b^{2}=c^{2}+2ab\cos\gamma$. >![[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.11.44.excalidraw.svg]] %%[[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.11.44.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 1/---files/Trigonometry_cosine_sine 2023-05-10 08.11.44.excalidraw.dark.svg|dark exported image]]%% --- ## 1 Some intuition. An angle is called acute if it is between $0^{\circ}$ and $90^{\circ}$, and an angle is called obtuse if it is between $90^{\circ}$ and $180^{\circ}$. (A) In a triangle, at most how many obtuse angles can you have? (B) In a triangle, if a side is the longest, what can you say about the angles it makes with the two shorter ones? (C) Suppose we have two acute angles \alpha and \beta that is between $0^{\circ}$ and $90^{\circ}$, with $\alpha < \beta$. How would $\sin\alpha$ and $\sin\beta$ compare? How would $\cos\alpha$ and $\cos\beta$ compare? Can you argue this by drawing two triangles with the same hypotenuse, but one with angle \alpha and the other with angle \beta? --- ## 2 Which angle is wider? In each of the following three triangles, the sides are given and an angle is marked. They are not drawn to scale. Which angle is the widest? And which one is the narrowest? --- ## 3 Proving law of cosine. Let us prove the law of cosine! Use the following diagram, and draw an altitude of length h perpendicular to side a. This splits a into two segments, a1 and a2. Can you show the law of cosine using Pythagorean theorem? Hint: Express h and a2 in terms of γ, and relate all the lengths that you know. ![[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.13.10.excalidraw.svg]] %%[[1 teaching/summer program 2023/week 2/---files/Trigonometry_cosine_sine 2023-05-10 08.13.10.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 1/---files/Trigonometry_cosine_sine 2023-05-10 08.13.10.excalidraw.dark.svg|dark exported image]]%% --- ## 4 Hero's formula One can actually express the area of a triangle by using just its side lengths. If a triangle has side lengths $a,b,c$, then its area is given by $A=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$. This is commonly phrased in this alternate way: Denote $s=\frac{1}{2}(a+b+c)$ be the **semiperimeter** of the triangle, then the area is given by $A=\sqrt{s(s-a)(s-b)(s-c)}$ This simple and curious looking formula is called Hero's formula (or Heron's). Try to show why this is true! Hint: Try to do the following (1) Draw an arbitrary triangle, and label the sides $a,b,c$. (2) Draw an **altitude** from one vertex to its opposing side, and write this **height** in terms of sine of some angle. (3) Use that angle to write a relation between the sides using law of cosine. (4) Write down the expression for the area of the triangle. At some point you will get some squares inside a square root. Don't expand it and instead factor using difference of squares, several times. You can do it! #summer-program-2023