A-level Mathematics/OCR/C2/Trigonometric Functions

The Law of Cosines
Pythagoras theory only applies to right triangles, the law of cosines will apply to any triangle. When you have a right triangle it reduces to the same formula as given by Pythagoras theorem. For any triangle ABC with angle measurement $$\alpha$$, $$\beta$$, $$\gamma$$ and sides of length a,b,c.

$$a^2=b^2 + c^2 - 2bc \cos \alpha \,$$ $$b^2=a^2 + c^2 - 2ac \cos \beta \,$$ $$c^2=a^2 + b^2 - 2ab \cos \gamma \,$$

Example

What is the value of c when a = 4 cm, b = 8 cm, and $$\gamma$$ is equal to $$64^\circ$$. $$c^2=4^2 + 8^2 - 2 \times 4 \times 8 \cos 64^\circ \,$$

$$c^2 = 53\,$$

$$c \approx 7.28\ cm$$

The Law of Sines
For any triangle ABC with angle measurement $$\alpha$$, $$\beta$$, $$\gamma$$ and sides of length a,b,c.

$$\frac {a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac {c}{\sin \gamma}$$

Example If Angle &alpha; is $$45^\circ$$, Angle &beta; is $$24^\circ$$ and Side b is 3 cm, what is the length of side a?

$$\frac {a}{\sin 45^\circ} = \frac{3}{\sin 24^\circ}$$

$$a \times \sin 24^\circ = 3 \times \sin 45^\circ$$

$$a= \frac{3 \times \sin 45^\circ}{\sin 24^\circ} \approx 5.22\ \mbox{cm}$$

Area of a Triangle
For any triangle the area is one-half the product of two sides with the sine of the included angle. If the included angle is a right angle, then this reduces to the formula for the area of a right triangle, since $$\sin 90^\circ= 1$$

$$Area = \frac{1}{2}bc \sin \alpha \,$$

$$Area = \frac{1}{2}ac \sin \beta \,$$

$$Area = \frac{1}{2}ab \sin \gamma \,$$

Example:

What is the area of triangle when a = 4 cm, b = 8 cm, and $$\gamma$$ is equal to $$20^\circ$$.

$$Area = \frac{1}{2}\times 4 \times 8 \times \sin 20^\circ\, \approx 5.47\ \mbox{cm}^2$$

Pythagoras Identity
$$\sin ^2 \theta + \cos ^2 \theta = 1 \,$$

Proof:

We use the pythagorean theory:

$$a^2 + b^2 = c^2\,$$

Now we divide by $$c^2$$:

$$\frac {a^2}{c^2} + \frac {b^2}{c^2} = \frac {c^2}{c^2}\,$$

We get:

$$\frac {a^2}{c^2} + \frac {b^2}{c^2} = 1\,$$

We can write this as:

$$\sin ^2 \theta + \cos ^2 \theta = 1 \,$$

A good way to think of this of is $$opposite^2 + adjacent^2 = hypotenuse^2 = \frac{opposite^2}{hypotenuse^2} + \frac{adjacent^2}{hypotenuse^2} = 1$$

A Practical Example
Find all the values of x between 0 rad and 2π rad  that satisfy the relationship $$15\sin^2\left(x\right) = \cos\left(x\right) + 13$$.

Using the Pythagoras Identity we get:

$$15\left(1 - \cos^2\left(x\right)\right) = \cos\left(x\right) + 13$$

Now we can simplify:

$$15\cos^2\left(x\right) - \cos\left(x\right) - 2 = 0$$

It is more covinent to replace cos(x) with u:

$$15u^2 - u - 2 = 0\,$$

Then we factor the expression

$$\left(5u + 2\right)\left(3u - 1\right) = 0$$

$$\cos\left(x\right) = \frac{-2}{5}\ or\ \frac{1}{3}$$

In order to determine what x is we need to use $$\cos^{-1}\left(x\right)$$ on our calculators.

$$\cos^{-1}\left(\frac{-2}{5}\right) \approx 1.9823\ rad$$

$$\cos^{-1}\left(\frac{1}{3}\right) \approx 1.2310\ rad$$

But we need to remember that in the interval 2π the cosine function will have the same in 2π - x.

2π rad - 1.2310 rad = 5.0222 rad

2π rad - 1.9823 rad = 4.3009 rad

So the complete answer is 1.2310 rad, 1.9823 rad, 4.3009 rad, and 5.0222 rad.

Tangent Identity
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

Proof:

$$\tan \theta = \frac{a}{b}$$

Then we can divide both the numerator and the denominator by c

$$\tan \theta = \frac {\frac {a}{c}}{ \frac {b}{c}}$$

We can write this as:

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

Example
sin(x) = 4cos(x) solve for sin(x). All units are in radians.

We divide both sides by cos x and we get the identity

tan(x)=4

We use the $$tan^{-1}(x) $$ to get that x = 1.3258 rad.

Now we can solve for sin(x):

sin(x) = 4cos(1.3258 rad) = 4*.2425 rad = .9701 rad.

Matemática elementar/Trigonometria/Funções trigonométricas