Solving trigonometric equations

December 23, 2016 Mouctar Trigonometry 0

Solving trigonometric equations

Angle of elevation:

In most trigonometric problems, you’ll see the word “Angle of elevation”.

The angle of elevation of an object is the angle formed by the horizontal and the line from that object with the vertex being the point of observation (eye of observer).

The observer looks up to see the object being observed.

In  this figure \angle A or \angle \alpha is the angle of elevation.

Angle of depression:

IThe angle of depression of an object is the angle formed by the horizontal and the line from that object with the vertex being the point of observation (eye of observer).

If you look at the figure, in this situation the observer is at a level higher that the one of the object. The observer looks down to see the object being observed.

In  this figure \angle \beta or \angle C is the angle of depression.

Right triangle Formulas

We have already seen these for the unit circle.

The challenge here is that our hypotenuse may never be 1.

We can,however,divide all the sides, including the hypotenuse, by the hypotenuse and we get:

For a non-right angle in the triangle:

\sin \alpha=opposite \; side for the unit circle.

\cos \alpha=adjacent \; side for the unit circle.

These two formulas become:

\sin \alpha=\frac{opposite \; side}{hypotenuse}

\cos \alpha=\frac{adjacent \; side}{hypotenuse}

The tangent remains unchanged:

\tan \alpha=\frac{opposite \; side}{adjacent \; side}

Most of the elementary trigonometry problems will be solved using these formulas.

We can see now why we should create a right angle for any given problem, where that can help, and solve using these simple formulas.

Inverse trigonometric functions:

Our trigonometric functions being one-to-one functions, we can find their inverse functions.

Those functions are of the form:

If y=f(x), we can reverse the correspondence and get x=f^{-1}(y).

We will have a special chapter on inverse functions.

For x=\sin y we have:

y=\sin^{-1}x for -1\leq x \leq 1 and -\frac{\pi}{2}\leq x \leq \frac{\pi}{2}

We have the property:

\sin (\sin^{-1}x)=\sin (\arcsin x)=x if  -1\leq x \leq 1

\sin^{-1}(\sin y)=\arcsin (\sin y)=y if -\frac{\pi}{2}\leq y \leq \frac{\pi}{2}

Example:

Find y for:

y=\sin^{-1}(\tan \frac{3\pi}{4})

\tan \frac{3\pi}{4}=-1

y=\sin^{-1} (-1)=-\frac{\pi}{2}

For x=\cos y we have:

y=\cos^{-1}x for -1\leq x \leq 1 and 0\leq x \leq \pi

Please read “y is the inverse cosine of x” or arccosine

We have the property:

\cos (\cos^{-1}x)=\cos (\arccos x)=x if  -1\leq x \leq 1

\cos^{-1}(\cos y)=\arccos (\cos y)=y if 0\leq y \leq \pi

For x=\tan y we have:

y=\tan^{-1}x=\arctan x iff x=\tan y

for any x and for -\frac{\pi}{2}\leq y \leq \frac{\pi}{2}

Please read “y is the inverse tangent of x” or arctangent

We have the property:

\tan (\tan^{-1}x)=\tan (\arctan x)=x  for every x

\tan^{-1}(\tan y)=\arctan (\tan y)=y if  -\frac{\pi}{2}\leq y \leq \frac{\pi}{2}

Example:

Find the value of:

y=\sin (\arctan \frac{1}{2}-\arccos \frac{4}{5})

if x_1=\arctan \frac{1}{2} we can have \tan x_1=\frac{1}{2}

The hypotenuse here is \sqrt{1^2+2^2}=\sqrt{5}.

Now :

\sin x_1=\frac{1}{\sqrt{5}}

\cos x_1=\frac{2}{\sqrt{5}}

if x_2=\arccos \frac{4}{5} we can have \cos x_2=\frac{4}{5}

The hypotenuse here is 5.

Now :

\sin x_2=\frac{3}{5}

All we need is \sin (x_1-x_2)

\sin (x_1-x_2)=\sin x_1 \cos x_2-\cos x_1 \sin x_2=\frac{1}{\sqrt{5}}\cdot \frac{4}{5}-\frac{2}{\sqrt{5}}\cdot \frac{3}{5}

After simplification we have:

y=\sin(\arctan \frac{1}{2}-\arccos \frac{4}{5})=-\frac{2\sqrt{5}}{25}

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