Question 28 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

cos⁡(3π4+x)−cos⁡(3π4−x)=−2 sin⁡x\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\,\sin x
Correct Answer

L.H.S. =cos⁡(3π4+x)−cos⁡(3π4−x)= \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right)

Using the identity cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B and cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B:

cos⁡(3π4+x)=cos⁡(3π4)cos⁡x−sin⁡(3π4)sin⁡x\cos\left(\frac{3\pi}{4} + x\right) = \cos\left(\frac{3\pi}{4}\right) \cos x - \sin\left(\frac{3\pi}{4}\right) \sin x cos⁡(3π4−x)=cos⁡(3π4)cos⁡x+sin⁡(3π4)sin⁡x\cos\left(\frac{3\pi}{4} - x\right) = \cos\left(\frac{3\pi}{4}\right) \cos x + \sin\left(\frac{3\pi}{4}\right) \sin x

Therefore,

cos⁡(3π4+x)−cos⁡(3π4−x)=[cos⁡(3π4)cos⁡x−sin⁡(3π4)sin⁡x]−[cos⁡(3π4)cos⁡x+sin⁡(3π4)sin⁡x]\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = \left[\cos\left(\frac{3\pi}{4}\right) \cos x - \sin\left(\frac{3\pi}{4}\right) \sin x\right] - \left[\cos\left(\frac{3\pi}{4}\right) \cos x + \sin\left(\frac{3\pi}{4}\right) \sin x\right] =cos⁡(3π4)cos⁡x−sin⁡(3π4)sin⁡x−cos⁡(3π4)cos⁡x−sin⁡(3π4)sin⁡x= \cos\left(\frac{3\pi}{4}\right) \cos x - \sin\left(\frac{3\pi}{4}\right) \sin x - \cos\left(\frac{3\pi}{4}\right) \cos x - \sin\left(\frac{3\pi}{4}\right) \sin x =−2sin⁡(3π4)sin⁡x= -2 \sin\left(\frac{3\pi}{4}\right) \sin x

Now, 3π4=π−π4\frac{3\pi}{4} = \pi - \frac{\pi}{4}, so

sin⁡(3π4)=sin⁡(π−π4)=sin⁡(π4)=12\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

Therefore,

L.H.S.=−2×12×sin⁡x=−2 sin⁡x=R.H.S.\text{L.H.S.} = -2 \times \frac{1}{\sqrt{2}} \times \sin x = -\sqrt{2}\,\sin x = \textbf{R.H.S.}

Hence proved.

Exercise: EXERCISE 3.3 | Q: 11 | (Page 25)
For More Understanding

Explanation

This question tests the application of cosine addition and subtraction formulas. The key insight is that when we subtract cos⁡(A−B)\cos(A-B) from cos⁡(A+B)\cos(A+B), the cos⁡Acos⁡B\cos A \cos B terms cancel out, leaving only −2sin⁡Asin⁡B-2 \sin A \sin B.

Students must also recall that sin⁡(3π4)=sin⁡(π4)=12\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} since 3π4\frac{3\pi}{4} lies in the second quadrant where sine is positive.

This follows the same pattern as Example 15 in the textbook which proves cos⁡(π4+x)+cos⁡(π4−x)=2 cos⁡x\cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) = \sqrt{2}\,\cos x.

Solution Steps

  1. Step 1: Apply cosine addition formula:

    cos⁡(3π4+x)=cos⁡(3π4)cos⁡x−sin⁡(3π4)sin⁡x\cos\left(\frac{3\pi}{4} + x\right) = \cos\left(\frac{3\pi}{4}\right) \cos x - \sin\left(\frac{3\pi}{4}\right) \sin x
  2. Step 2: Apply cosine subtraction formula:

    cos⁡(3π4−x)=cos⁡(3π4)cos⁡x+sin⁡(3π4)sin⁡x\cos\left(\frac{3\pi}{4} - x\right) = \cos\left(\frac{3\pi}{4}\right) \cos x + \sin\left(\frac{3\pi}{4}\right) \sin x
  3. Step 3: Subtract the two expressions: cos⁡(3π4)cos⁡x\cos\left(\frac{3\pi}{4}\right) \cos x terms cancel, leaving −2sin⁡(3π4)sin⁡x-2 \sin\left(\frac{3\pi}{4}\right) \sin x

  4. Step 4: Evaluate sin⁡(3π4)=sin⁡(π−π4)=sin⁡(π4)=12\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

  5. Step 5: Simplify:

    −2×12×sin⁡x=−2 sin⁡x-2 \times \frac{1}{\sqrt{2}} \times \sin x = -\sqrt{2}\,\sin x