Trigonometric Identities

**jonathanraxa** · April 10th 2012, 06:06 PM

Show that:

sin²(angle) = [tan²(angle)]/[1+tan²(angle)]

fir all (angles) except odd multiples of pi/2.

(This is coming straight from my homework)

-Thank you!

**Soroban** · April 10th 2012, 07:51 PM

Hello, jonathanraxa!

$\text{Show that: }\:\sin^2\!x \:=\:\frac{\tan^2\!x}{1+\tan^2x}$

. . . $\text{for all angles except odd multiples of }\tfrac{\pi}{2}$

The right side is:
. . $\frac{\tan^2\!x}{1+\tan^2\!x} \;\;=\;\;\frac{\tan^2\!x}{\sec^2\!x} \;\;=\;\;\frac{\dfrac{\sin^2\!x}{\cos^2\!x}}{ \dfrac{1}{\cos^2\!x}} \;\;=\;\;\frac{\sin^2\!x}{\cos^2\!x}\cdot \frac{\cos^2\!x}{1} \;\;=\;\;\sin^2\!x$

**Prove It** · April 10th 2012, 07:52 PM

Originally Posted by jonathanraxa

Show that:

sin²(angle) = [tan²(angle)]/[1+tan²(angle)]

fir all (angles) except odd multiples of pi/2.

(This is coming straight from my homework)

-Thank you!

First note that $\displaystyle \begin{align*} \tan{\theta} \end{align*}$ is undefined for all odd multiples of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ (do you know why?) so that means that any combination of $\displaystyle \begin{align*} \tan{\theta} \end{align*}$ will also be undefined at all odd multiples of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ .

$\displaystyle \begin{align*} \frac{\tan^2{\theta}}{1 + \tan^2{\theta}} &\equiv \frac{\frac{\sin^2{\theta}}{\cos^2{\theta}}}{1 + \frac{\sin^2{\theta}}{\cos^2{\theta}}} \\ &\equiv \frac{\frac{\sin^2{\theta}}{\cos^2{\theta}}}{\frac {\cos^2{\theta} + \sin^2{\theta}}{\cos^2{\theta}}} \\ &\equiv \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} \left(\cos^2{\theta} + \sin^2{\theta} \right)} \\ &\equiv \frac{\sin^2{\theta}}{\cos^2{\theta} + \sin^2{\theta}} \\ &\equiv \frac{\sin^2{\theta}}{1} \\ &\equiv \sin^2{\theta} \end{align*}$

**highvoltage** · April 10th 2012, 08:07 PM

Thread: Trigonometric Identities

LinkBack

Thread Tools

Display