# How do you use the fundamental trigonometric identities to determine the simplified form of the expression?

Aug 8, 2016

"The fundamental trigonometric identities" are the basic identities:

•The reciprocal identities
•The pythagorean identities
•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!