The identity is proved by expressing each product as a sum using the cosine product-to-sum formula, simplifying the resulting sum of cosines, and then converting back to a product using the sine sum-to-product formula. The final result is sin5xsin25x.
Concept and Intuition
When you see a trigonometric identity involving products of cosines on one side and a product of sines on the other, the natural instinct is to reach for the product-to-sum formulas. These formulas turn products into sums, which are much easier to combine and simplify. After simplifying the sum, you can often reverse the process using a sum-to-product formula to get back a single product — exactly what we need here.
The key insight: both terms on the left are of the form cosAcosB. Using the identity
cosAcosB=21[cos(A+B)+cos(A−B)],
we can rewrite each term, then add them, and watch the cancellations happen. The final sum will collapse into a single difference of cosines, which we then rewrite as a product of sines.
Step-by-step Proof
1. Apply the product-to-sum formula to the first term.
We have cos2xcos2x. Using
cosAcosB=21[cos(A+B)+cos(A−B)],
with A=2x and B=2x, we get:
cos2xcos2x=21[cos(2x+2x)+cos(2x−2x)]
Simplify the arguments:
2x+2x=24x+2x=25x,2x−2x=24x−2x=23x
So:
cos2xcos2x=21[cos25x+cos23x]
2. Apply the product-to-sum formula to the second term.
Now cos3xcos29x. Here A=3x, B=29x.
cos3xcos29x=21[cos(3x+29x)+cos(3x−29x)]
Simplify:
3x+29x=26x+29x=215x,3x−29x=26x−29x=−23x
Since cos(−θ)=cosθ, we have cos(−23x)=cos23x. Thus:
cos3xcos29x=21[cos215x+cos23x]
3. Subtract the two expressions.
The left-hand side of the identity is cos2xcos2x−cos3xcos29x. Substituting:
LHS=21[cos25x+cos23x]−21[cos215x+cos23x]
Factor out 21:
=21[cos25x+cos23x−cos215x−cos23x]
Notice cos23x cancels out:
=21[cos25x−cos215x]
4. Use the sum-to-product formula for a difference of cosines.
The formula is:
cosP−cosQ=−2sin2P+Qsin2P−Q
Here P=25x, Q=215x. Compute:
2P+Q=225x+215x=2220x=210x=5x
2P−Q=225x−215x=2−210x=2−5x
Thus:
cos25x−cos215x=−2sin(5x)sin(−25x)
Since sin(−θ)=−sinθ, we have sin(−25x)=−sin25x. So:
=−2sin(5x)⋅(−sin25x)=2sin5xsin25x
5. Substitute back into the expression.
We had LHS=21[cos25x−cos215x]. Replacing the bracket:
LHS=21⋅2sin5xsin25x=sin5xsin25x
This is exactly the right-hand side.
A common mistake is forgetting the negative sign in the cosine difference formula, or mishandling the sign when sin(−θ) appears. Always check: cosP−cosQ=−2sin2P+Qsin2P−Q, and sin(−θ)=−sinθ. The two negatives multiply to give a positive.
If you ever get stuck, try working backwards from the RHS using the product-to-sum formula for sines: sin5xsin25x=21[cos(5x−25x)−cos(5x+25x)]=21[cos25x−cos215x], which is exactly what we obtained in step 3. This confirms the path is correct.
✓Final answer
The identity is proved: cos2xcos2x−cos3xcos29x=sin5xsin25x.