How to prove this question with using a LS=RS-style Proof?
csc^2(x)cos^2(x) = csc^2(x)-1
This type of problem involves a lot of trig idenities, I just looked them up..
Lets look at the LHS:
csc^2(x)= 1/sin^2(x)
cos^2(x) = 1-sin^2(x)
Now we can say:
csc^2(x)*cos^2(x) = (1-sin^2(x))/sin^2(x)
now lets look at the RHS:
csc^2(x) = 1/sin^2(x)
and we can change that 1 to sin^2(x)/sin^2(x) to have same denominator
1/sin^2(x) - sin^2(x)/sin^2(x) = (1-sin^2(x))/sin^2(x)
So now LHS=RHS
(1-sin^2(x))/sin^2(x) = (1-sin^2(x))/sin^2(x)
From the identity sin^2 x + cos^2 x = 1
if we divide both sides by sin^2 x , we get
1 + cot^2 x = csc^2 x
so csc^2 x - 1 = cot^2 x
RS = cot^2 x
LS = csc^2(x)cos^2(x)
= (1/sin^2 x)(cos^2 x)
= cot^2 x
= RS
All done
@Don & Reiny: thank you very much..
I really understand your answers after 3 hours of tried them by myself. thank you so much.. :D
To prove the equation csc^2(x) * cos^2(x) = csc^2(x) - 1 using a LS=RS (left side equals right side) proof, we need to simplify both sides of the equation and show that they are equal.
Starting with the left side (LS):
LS = csc^2(x) * cos^2(x)
Now, let's simplify the left side by using the trigonometric identity: sin^2(x) + cos^2(x) = 1. Rearranging this identity, we get: cos^2(x) = 1 - sin^2(x).
Substituting this into the left side equation, we have:
LS = csc^2(x) * (1 - sin^2(x))
Next, we need to simplify the right side (RS):
RS = csc^2(x) - 1
Now, let's simplify the right side by using the reciprocal identity: csc(x) = 1/sin(x). Squaring both sides of this identity, we get: csc^2(x) = 1/sin^2(x).
Substituting this into the right side equation, we have:
RS = 1/sin^2(x) - 1
To make the denominators the same, we will multiply the second term on the right side by sin^2(x)/sin^2(x):
RS = 1/sin^2(x) - sin^2(x)/sin^2(x)
Combining the terms:
RS = (1 - sin^2(x))/sin^2(x)
Now, we can see that both the left side (LS) and the right side (RS) are in a similar form. If we can show that they are equal, the equation will be proved.
To simplify both sides further, let's use the pythagorean identity: sin^2(x) + cos^2(x) = 1. Rearranging it, we get: sin^2(x) = 1 - cos^2(x).
Substituting this into the right side equation:
RS = (1 - (1 - cos^2(x)))/sin^2(x)
RS = (1 - 1 + cos^2(x))/sin^2(x)
RS = cos^2(x)/sin^2(x)
Now, let's simplify the left side:
LS = csc^2(x) * (1 - sin^2(x))
LS = (1/sin^2(x)) * (1 - (1 - cos^2(x)))
LS = (1/sin^2(x)) * (cos^2(x))
LS = cos^2(x)/sin^2(x)
As you can see, the left side (LS) and right side (RS) are equal. Therefore, we have proven the equation: csc^2(x) * cos^2(x) = csc^2(x) - 1 using a LS=RS-style proof.