Error in the proof starts with the first mistake done. The erroneous step in the proof is given by: Option B: Step 6.
Sign changing in equation does by multiplying -1 on both the sides.
So, if we had [tex]a + b = c[/tex]
Then, its signs are changed as:
[tex]-1(a + b) = -1 \times c\\-a -b = -c[/tex]
For the given case, the mistake occurred during transition from step 5 to step 6. Step 6 is wrong as when sign was changed, sin of cos(θ) remain same, which makes step 5 and step 6 not equal to each other.
The correct proof would be like:
[tex]\rm \cos(2x) = 1- 2\sin^2(x)\\Let\: 2x = \theta\\\\Then: x = \dfrac{\theta}{2}\\\\\cos(\theta) = 1 - 2\sin^2(\dfrac{\theta}{2})\\-1 + \cos(\theta) = -2\sin^2(\dfrac{\theta}{2})\\\\1 - \cos(\theta) = 2\sin^2(\dfrac{\theta}{2})\\\\\dfrac{1-\cos(\theta)}{2} = \sin^2(\dfrac{\theta}{2})\\\\\sin(\theta) = \pm \sqrt{\dfrac{1-\cos(\theta)}{2} }[/tex]
Thus, the erroneous step in the proof is given by: Option B: Step 6.
Learn more about trigonometric functions here:
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