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trig identities
- Thread starter dcgirl
- Start date
G'day, Dcgirl.
Begin with the more complicated side, here the LHS.
. . \(\displaystyle \L LHS \, = \, \cos{x} \, + \, \sin{x}\tan{x}\)
Simplify \(\displaystyle \L \sin{x}\tan{x}\):
. . \(\displaystyle \L \sin{x}\tan{x} \, = \, \sin{x} \, \cdot \, \frac{\sin{x}}{cos{x}} \, = \, \frac{sin^2{x}}{\cos{x}}\)
We now have
. . \(\displaystyle \L LHS \, = \, \cos{x} \, + \, \frac{sin^2{x}}{\cos{x}}\).
Treat as
. . \(\displaystyle \L LHS \, = \, \frac{\cos{x}}{1} \, + \, \frac{sin^2{x}}{\cos{x}}\)
And combine the fractions (find the lowest common denominator/cross multiply). You should recognise the identity in the resultant numerator, to show LHS = RHS.
Begin with the more complicated side, here the LHS.
. . \(\displaystyle \L LHS \, = \, \cos{x} \, + \, \sin{x}\tan{x}\)
Simplify \(\displaystyle \L \sin{x}\tan{x}\):
. . \(\displaystyle \L \sin{x}\tan{x} \, = \, \sin{x} \, \cdot \, \frac{\sin{x}}{cos{x}} \, = \, \frac{sin^2{x}}{\cos{x}}\)
We now have
. . \(\displaystyle \L LHS \, = \, \cos{x} \, + \, \frac{sin^2{x}}{\cos{x}}\).
Treat as
. . \(\displaystyle \L LHS \, = \, \frac{\cos{x}}{1} \, + \, \frac{sin^2{x}}{\cos{x}}\)
And combine the fractions (find the lowest common denominator/cross multiply). You should recognise the identity in the resultant numerator, to show LHS = RHS.
I find it's good guide to decide which side looks the most complicated, and look to simplify it down to the less-complicated side.
Identities do become more complicated, though, but it's just a matter of chugging through the exercises and you'll find your foresight improves greatly.
Identities do become more complicated, though, but it's just a matter of chugging through the exercises and you'll find your foresight improves greatly.