Give an example, with justification:

A real life situation requiring the use of a trigonometry identity.

Knowledge is life, so open your doors of perception.

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- Aug 23rd 2008, 07:15 AMpuneettrigonometry
Give an example, with justification:

A real life situation requiring the use of a trigonometry identity.

Knowledge is life, so open your doors of perception. - Aug 23rd 2008, 08:33 AMTKHunny
Anything you want square. Measuring the sides will not get you there unless:

1) You can verify the corners are right angles, or

2) You can calculate and measure the diagonals. - Aug 23rd 2008, 08:35 AMskeeter
Here is one use of identities that is introduced to students in a high school physics class ... the derivation of the range formula for the horizontal displacement of a projectile (Galilean model).

start with the physics equation for vertical displacement ...

$\displaystyle \Delta y = v_0 \sin(\theta) t - \frac{1}{2}gt^2$

since the projectile lands at the same height it started,

$\displaystyle \Delta y = 0$ ...

$\displaystyle 0 = v_0 \sin(\theta) t - \frac{1}{2}gt^2$

solving for t ...

$\displaystyle t = \frac{2v_0 \sin(\theta)}{g}$

horizontal displacement is given by the formula ...

$\displaystyle \Delta x = v_0 \cos(\theta) t$

using the last two equations, eliminate the parameter t ...

$\displaystyle \Delta x = \frac{2v_0^2 \sin(\theta) \cos(\theta)}{g}$

**using the double angle identity**$\displaystyle \sin(2\theta) = 2\sin(\theta)\cos(\theta)$,

the last equation becomes ...

$\displaystyle \Delta x = \frac{v_0^2 \sin(2\theta)}{g}$

from this equation, it can be deduced that the launch angle for maximum horizontal range is 45 degrees.