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Miscellaneous Examples · Example 19

Q.Prove that cos⁡2x cos⁡x2−cos⁡3x cos⁡9x2=sin⁡5x sin⁡5x2\cos 2x\, \cos\frac{x}{2} - \cos 3x\, \cos\frac{9x}{2} = \sin 5x\, \sin\frac{5x}{2}.

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✓ Free question

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 sin⁡5xsin⁡5x2\sin 5x \sin\frac{5x}{2}.

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 cos⁡Acos⁡B\cos A \cos B. Using the identity

cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)]\cos A \cos B = \frac{1}{2}[\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 cos⁡2xcos⁡x2\cos 2x \cos\frac{x}{2}. Using

cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)]\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)],

with A=2xA = 2x and B=x2B = \frac{x}{2}, we get:

cos⁡2xcos⁡x2=12[cos⁡(2x+x2)+cos⁡(2x−x2)]\cos 2x \cos\frac{x}{2} = \frac{1}{2}\left[\cos\left(2x + \frac{x}{2}\right) + \cos\left(2x - \frac{x}{2}\right)\right]

Simplify the arguments:

2x+x2=4x2+x2=5x2,2x−x2=4x2−x2=3x22x + \frac{x}{2} = \frac{4x}{2} + \frac{x}{2} = \frac{5x}{2}, \quad 2x - \frac{x}{2} = \frac{4x}{2} - \frac{x}{2} = \frac{3x}{2}

So:

cos⁡2xcos⁡x2=12[cos⁡5x2+cos⁡3x2]\cos 2x \cos\frac{x}{2} = \frac{1}{2}\left[\cos\frac{5x}{2} + \cos\frac{3x}{2}\right]

2. Apply the product-to-sum formula to the second term.

Now cos⁡3xcos⁡9x2\cos 3x \cos\frac{9x}{2}. Here A=3xA = 3x, B=9x2B = \frac{9x}{2}.

cos⁡3xcos⁡9x2=12[cos⁡(3x+9x2)+cos⁡(3x−9x2)]\cos 3x \cos\frac{9x}{2} = \frac{1}{2}\left[\cos\left(3x + \frac{9x}{2}\right) + \cos\left(3x - \frac{9x}{2}\right)\right]

Simplify:

3x+9x2=6x2+9x2=15x2,3x−9x2=6x2−9x2=−3x23x + \frac{9x}{2} = \frac{6x}{2} + \frac{9x}{2} = \frac{15x}{2}, \quad 3x - \frac{9x}{2} = \frac{6x}{2} - \frac{9x}{2} = -\frac{3x}{2}

Since cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta, we have cos⁡(−3x2)=cos⁡3x2\cos\left(-\frac{3x}{2}\right) = \cos\frac{3x}{2}. Thus:

cos⁡3xcos⁡9x2=12[cos⁡15x2+cos⁡3x2]\cos 3x \cos\frac{9x}{2} = \frac{1}{2}\left[\cos\frac{15x}{2} + \cos\frac{3x}{2}\right]

3. Subtract the two expressions.

The left-hand side of the identity is cos⁡2xcos⁡x2−cos⁡3xcos⁡9x2\cos 2x \cos\frac{x}{2} - \cos 3x \cos\frac{9x}{2}. Substituting:

LHS=12[cos⁡5x2+cos⁡3x2]−12[cos⁡15x2+cos⁡3x2]\text{LHS} = \frac{1}{2}\left[\cos\frac{5x}{2} + \cos\frac{3x}{2}\right] - \frac{1}{2}\left[\cos\frac{15x}{2} + \cos\frac{3x}{2}\right]

Factor out 12\frac{1}{2}:

=12[cos⁡5x2+cos⁡3x2−cos⁡15x2−cos⁡3x2]= \frac{1}{2}\left[\cos\frac{5x}{2} + \cos\frac{3x}{2} - \cos\frac{15x}{2} - \cos\frac{3x}{2}\right]

Notice cos⁡3x2\cos\frac{3x}{2} cancels out:

=12[cos⁡5x2−cos⁡15x2]= \frac{1}{2}\left[\cos\frac{5x}{2} - \cos\frac{15x}{2}\right]

4. Use the sum-to-product formula for a difference of cosines.

The formula is:

cos⁡P−cos⁡Q=−2sin⁡P+Q2sin⁡P−Q2\cos P - \cos Q = -2 \sin\frac{P+Q}{2} \sin\frac{P-Q}{2}

Here P=5x2P = \frac{5x}{2}, Q=15x2Q = \frac{15x}{2}. Compute:

P+Q2=5x2+15x22=20x22=10x2=5x\frac{P+Q}{2} = \frac{\frac{5x}{2} + \frac{15x}{2}}{2} = \frac{\frac{20x}{2}}{2} = \frac{10x}{2} = 5x

P−Q2=5x2−15x22=−10x22=−5x2\frac{P-Q}{2} = \frac{\frac{5x}{2} - \frac{15x}{2}}{2} = \frac{-\frac{10x}{2}}{2} = \frac{-5x}{2}

Thus:

cos⁡5x2−cos⁡15x2=−2sin⁡(5x)sin⁡(−5x2)\cos\frac{5x}{2} - \cos\frac{15x}{2} = -2 \sin(5x) \sin\left(-\frac{5x}{2}\right)

Since sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta, we have sin⁡(−5x2)=−sin⁡5x2\sin\left(-\frac{5x}{2}\right) = -\sin\frac{5x}{2}. So:

=−2sin⁡(5x)⋅(−sin⁡5x2)=2sin⁡5xsin⁡5x2= -2 \sin(5x) \cdot \left(-\sin\frac{5x}{2}\right) = 2 \sin 5x \sin\frac{5x}{2}

5. Substitute back into the expression.

We had LHS=12[cos⁡5x2−cos⁡15x2]\text{LHS} = \frac{1}{2}\left[\cos\frac{5x}{2} - \cos\frac{15x}{2}\right]. Replacing the bracket:

LHS=12⋅2sin⁡5xsin⁡5x2=sin⁡5xsin⁡5x2\text{LHS} = \frac{1}{2} \cdot 2 \sin 5x \sin\frac{5x}{2} = \sin 5x \sin\frac{5x}{2}

This is exactly the right-hand side.

Watch out

A common mistake is forgetting the negative sign in the cosine difference formula, or mishandling the sign when sin⁡(−θ)\sin(-\theta) appears. Always check: cos⁡P−cos⁡Q=−2sin⁡P+Q2sin⁡P−Q2\cos P - \cos Q = -2 \sin\frac{P+Q}{2} \sin\frac{P-Q}{2}, and sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta. The two negatives multiply to give a positive.

Tip

If you ever get stuck, try working backwards from the RHS using the product-to-sum formula for sines: sin⁡5xsin⁡5x2=12[cos⁡(5x−5x2)−cos⁡(5x+5x2)]=12[cos⁡5x2−cos⁡15x2]\sin 5x \sin\frac{5x}{2} = \frac{1}{2}[\cos(5x - \frac{5x}{2}) - \cos(5x + \frac{5x}{2})] = \frac{1}{2}[\cos\frac{5x}{2} - \cos\frac{15x}{2}], which is exactly what we obtained in step 3. This confirms the path is correct.


✓Final answer

The identity is proved: cos⁡2xcos⁡x2−cos⁡3xcos⁡9x2=sin⁡5xsin⁡5x2\cos 2x \cos\frac{x}{2} - \cos 3x \cos\frac{9x}{2} = \sin 5x \sin\frac{5x}{2}.

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