Proof of Trigonometry Identity: a^2b^2=sin^2θcos^2θ

But I'll write my solution down anyway for the sake of completeness:In summary, we have two equations: ##a^2= \frac {sin^4θ} {cos^2θ}## and ##b^2=\frac {cos^4θ} {sin^2θ}##. Multiplying these two equations, we get ##a^2b^2= sin^2θ. cos^2θ##. We also have the equation ##a^2+b^2+3= \frac {sin^6θ + cos^6θ + 3 sin^2θ cos^2θ} {sin^2θ cos^θ}##. Multiplying ##a^2b^2
  • #1
chwala
Gold Member
2,692
354
Homework Statement
if ##secθ - cosθ =a##
## cosec θ-sin θ = b## then prove that
##a^2b^2(a^2+b^2+3) =1##
Relevant Equations
trigonometry
##a^2= \frac {sin^4θ} {cos^2θ}##
##b^2=\frac {cos^4θ} {sin^2θ}##
therefore
##a^2b^2= sin^2θ. cos^2θ##...1
and
##a^2+b^2+3= \frac {sin^6θ + cos^6θ + 3 sin^2θ cos^2θ} {sin^2θ cos^θ}##...2
multiplying 1 and 2, we get,
##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## how do i proceed from here?

i think i got it
##sin^6θ + cos^6θ =(sin^2x)^3 + (cos^2x)^3##
using,
##a^3 + b^3 = (a+b)(a^2+b^2-ab)##
= ##(sin^2x + cos^2x)(sin^4x + cos^4x-sin^2x. cos^2x)##
and using,
##a^2+b^2= (a+b)^2-2ab##
##(a^2+b^2-ab)= (a+b)^2-3ab##
thus,
=## (sin^2x + cos^2x)^2-3sin^2x. cos^2x##
therefore,

##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## = ## [(sin^2x + cos^2x)^2-3sin^2x. cos^2x]+3 sin^2θ cos^2θ##
=1
Bingo!any alternative method guys?
 
Last edited:
Physics news on Phys.org
  • #2
I did it this way, writing s and c for sin and cos.
##ac=s^2##
##bs=c^2##
##a^2+b^2=\frac{s^4}{c^2}+\frac{c^4}{s^2}=\frac{s^6+c^6}{s^2c^2}##
Factorising sum of two cubes..
##=(s^2+c^2)\frac{s^4-s^2c^2+c^4}{s^2c^2}##
##=\frac{(s^2+c^2)^2-3s^2c^2}{s^2c^2}##
So ##a^2+b^2+3=\frac{1}{s^2c^2}##
And
##a^2b^2=s^2c^2##

Having gone through that, I see what you did is essentially the same.
 
Last edited:
  • Love
Likes etotheipi

Related to Proof of Trigonometry Identity: a^2b^2=sin^2θcos^2θ

1. What is the proof of the trigonometry identity a^2b^2 = sin^2θcos^2θ?

The proof of this identity is based on the Pythagorean identity for sine and cosine, which states that sin^2θ + cos^2θ = 1. By rearranging this identity, we can get sin^2θ = 1 - cos^2θ and cos^2θ = 1 - sin^2θ. Substituting these values into the original identity, we get a^2b^2 = (1 - cos^2θ)(1 - sin^2θ). Using the FOIL method, we can expand this to a^2b^2 = 1 - sin^2θ - cos^2θ + sin^2θcos^2θ. Simplifying this further, we get a^2b^2 = sin^2θcos^2θ, proving the identity.

2. How is this identity useful in trigonometry?

This identity is useful in solving trigonometric equations and proving other identities. It can also be used to simplify complex trigonometric expressions and make them easier to work with.

3. Can this identity be applied to any triangle?

Yes, this identity can be applied to any triangle, regardless of its size or shape. This is because the Pythagorean identity for sine and cosine holds true for all triangles.

4. Are there any other ways to prove this identity?

Yes, there are other ways to prove this identity, such as using the double angle formulas for sine and cosine or using the sum and difference formulas for sine and cosine. However, the method described in the first question is the most straightforward and commonly used method.

5. Can this identity be extended to more than two trigonometric functions?

Yes, this identity can be extended to more than two trigonometric functions. For example, we can prove that a^2b^2c^2 = sin^2θcos^2θtan^2θ using similar methods. However, the proof becomes more complex as the number of functions increases.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
969
  • Calculus and Beyond Homework Help
Replies
9
Views
503
  • Calculus and Beyond Homework Help
Replies
3
Views
452
Replies
1
Views
845
  • Precalculus Mathematics Homework Help
Replies
7
Views
733
  • Precalculus Mathematics Homework Help
Replies
7
Views
956
  • Calculus and Beyond Homework Help
Replies
8
Views
956
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
Back
Top