# Fundamental Identities

sin(A+B)=sin(A)cos(B)+cos(A)sin(B) proof - geometrical

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Tiago Hands

## Key Questions

• Those are formulas that you can derive from the pitagoras triangle.
For example:
${c}^{2} = {a}^{2} + {b}^{2}$
Expressed in sin and cos you have:
$1 = {\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)$
This is one identity, then you play a bit with this formula and get another identity.

Divide by ${\cos}^{2} \left(\theta\right)$:

$\frac{1}{{\cos}^{2} \left(\theta\right)} - \frac{{\cos}^{2} \left(\theta\right)}{{\cos}^{2} \left(\theta\right)} = \frac{{\sin}^{2} \left(\theta\right)}{{\cos}^{2} \left(\theta\right)}$

${\sec}^{2} \left(\theta\right) - 1 = t {g}^{2} \left(\theta\right)$

And keep going.

• "The fundamental trigonometric identities" are the basic identities:

â€¢The reciprocal identities
â€¢The pythagorean identities
â€¢The quotient identities

They are all shown in the following image:

When it comes down to simplifying with these identities, we must use combinations of these identities to reduce a much more complex expression to its simplest form.

Here are a few examples I have prepared:

a) Simplify: $\tan \frac{x}{\csc} x \times \sec x$

Apply the quotient identity $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and the reciprocal identities $\csc \theta = \frac{1}{\sin} \theta$ and $\sec \theta = \frac{1}{\cos} \theta$.

$= \frac{\sin \frac{x}{\cos} x}{\frac{1}{\sin} x} \times \frac{1}{\cos} x$

$= \sin \frac{x}{\cos} x \times \sin \frac{x}{1} \times \frac{1}{\cos} x$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x$

Reapplying the quotient identity, in reverse form:

$= {\tan}^{2} x$

b) Simplify: $\frac{\csc \beta - \sin \beta}{\csc} \beta$

Apply the reciprocal identity $\csc \beta = \frac{1}{\sin} \beta$:

$= \frac{\frac{1}{\sin} \beta - \sin \beta}{\frac{1}{\sin} \beta}$

Put the denominator on a common denominator:

$= \frac{\frac{1}{\sin} \beta - {\sin}^{2} \frac{\beta}{\sin} \beta}{\frac{1}{\sin} \beta}$

Rearrange the pythagorean identity ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$, solving for ${\cos}^{2} \theta$:

${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

$= \frac{{\cos}^{2} \frac{\beta}{\sin} \beta}{\frac{1}{\sin} \beta}$

$= {\cos}^{2} \frac{\beta}{\sin} \beta \times \sin \frac{\beta}{1}$

$= {\cos}^{2} \beta$

c) Simplify: $\sin \frac{x}{\cos} x + \cos \frac{x}{1 + \sin x}$:

Once again, put on a common denominator:

$= \frac{\sin x \left(1 + \sin x\right)}{\cos x \left(1 + \sin x\right)} + \frac{\cos x \left(\cos x\right)}{\cos x \left(1 + \sin x\right)}$

Multiply out:

$= \frac{\sin x + {\sin}^{2} x + {\cos}^{2} x}{\cos x \left(1 + \sin x\right)}$

Applying the pythagorean identity ${\cos}^{2} x + {\sin}^{2} x = 1$:

$= \frac{\sin x + 1}{\cos x \left(1 + \sin x\right)}$

Cancelling out the $\sin x + 1$ since it appears both in the numerator and in the denominator.

=cancel(sinx + 1)/(cosx(cancel(sinx + 1))

$= \frac{1}{\cos} x$

Applying the reciprocal identity $\frac{1}{\cos} \theta = \sec \theta$

$= \sec x$

Finally, on a last note, I know that here in Canada, British Columbia more specifically, these identities are given on a formula sheet, but I don't know what it's like elsewhere. In any event, many students, me included, memorize these identities because they're that important to mathematics. I would highly recommend memorization.

Practice exercises:

Simplify the following expressions:

a) $\cos \alpha + \tan \alpha \sin \alpha$

b) $\csc \frac{x}{\sin} x - \cot \frac{x}{\tan} x$

c) ${\sin}^{4} \theta - {\cos}^{4} \theta$

d) $\frac{\tan \beta + \cot \beta}{\csc} ^ 2 \beta$

Hopefully this helps, and good luck!

• Even & Odd Functions

A function $f \left(x\right)$ is said to be $\left\{\begin{matrix}\text{even if "f(-x)=f(x) \\ "odd if } f \left(- x\right) = - f \left(x\right)\end{matrix}\right.$

Note that the graph of an even function is symmetric about the $y$-axis, and the graph of an odd function is symmetric about the origin.

Examples

$f \left(x\right) = {x}^{4} + 3 {x}^{2} + 5$ is an even function since

$f \left(- x\right) = {\left(- x\right)}^{4} + {\left(- x\right)}^{2} + 5 = {x}^{4} + 3 {x}^{2} + 5 = f \left(x\right)$

$g \left(x\right) = {x}^{5} - {x}^{3} + 2 x$ is an odd function since

$g \left(- x\right) = {\left(- x\right)}^{5} - {\left(- x\right)}^{3} + 2 \left(- x\right) = - {x}^{5} + {x}^{3} - 2 x = - f \left(x\right)$

I hope that this was helpful.

• Divide the fundamental identity ${\sin}^{2} x + {\cos}^{2} x = 1$ by ${\sin}^{2} x$ or ${\cos}^{2} x$ to derive the other two:

${\sin}^{2} \frac{x}{\sin} ^ 2 x + {\cos}^{2} \frac{x}{\sin} ^ 2 x = \frac{1}{\sin} ^ 2 x$

$1 + {\cot}^{2} x = {\csc}^{2} x$

${\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$

${\tan}^{2} x + 1 = {\sec}^{2} x$

## Questions

• · 4 days ago
• · 6 days ago
• · 1 week ago
• · 2 weeks ago
• · 2 weeks ago
• · 3 weeks ago
• · 1 month ago
• · 1 month ago
• · 2 months ago
• 2 months ago
• · 2 months ago
• 3 months ago
• · 3 months ago
• · 4 months ago
• · 6 months ago
• · 6 months ago
• · 6 months ago
• · 6 months ago
• · 6 months ago
• · 7 months ago
• · 7 months ago
• · 7 months ago
• · 8 months ago
• · 8 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago